[proofplan] We prove the four properties in order. Property (i) bounds $N = \binom{2n}{n}$ between $2^{2n}/(2n)$ and $2^{2n}$ via two ways of comparing $N$ to the binomial sum $\sum_m \binom{2n}{m} = 2^{2n}$: once using that $N$ is one term, once using that $N$ is the maximum term. Properties (ii) and (iii) are prime-counting estimates on the factorisation $N = (2n)! / (n!)^2$: a prime $p \in (n, 2n]$ appears once in the numerator and not in the denominator, while a prime $p \in (2n/3, n]$ appears with the same multiplicity in both. Property (iv) is Legendre's prime-power counting formula $\nu_p(m!) = \sum_i \lfloor m/p^i \rfloor$ combined with the elementary fact that $\lfloor 2x \rfloor - 2\lfloor x \rfloor \in \{0, 1\}$. [/proofplan]

[step:Prove (i) by comparing $N$ to the binomial sum $\sum_m \binom{2n}{m} = 2^{2n}$] By the binomial theorem applied to $(1 + 1)^{2n}$, \begin{align*} \sum_{m=0}^{2n} \binom{2n}{m} = 2^{2n}. \end{align*} For the upper bound: since every binomial coefficient is non-negative and $N = \binom{2n}{n}$ is one of the $2n + 1$ terms, \begin{align*} N \leq \sum_{m=0}^{2n} \binom{2n}{m} = 2^{2n}. \end{align*} For the lower bound: we use the fact that $N = \binom{2n}{n}$ is the maximum of the binomial coefficients $\binom{2n}{m}$ for $0 \leq m \leq 2n$. Indeed, the ratio $\binom{2n}{m+1} / \binom{2n}{m} = (2n - m) / (m + 1)$ exceeds $1$ when $m < n$ and is less than $1$ when $m \geq n$, so the sequence $m \mapsto \binom{2n}{m}$ is strictly increasing on $\{0, 1, \ldots, n\}$ and strictly decreasing on $\{n, n+1, \ldots, 2n\}$, attaining its maximum at $m = n$. Separating the two extreme terms $\binom{2n}{0} = \binom{2n}{2n} = 1$ from the remaining $2n - 1$ terms, each bounded above by $N$, \begin{align*} 2^{2n} = \sum_{m=0}^{2n} \binom{2n}{m} \leq 2 + (2n - 1) N \leq 2n \cdot N, \end{align*} where in the last inequality we used $2 \leq N$ (valid for $n \geq 1$, since $N = \binom{2n}{n} \geq \binom{2}{1} = 2$). Rearranging yields \begin{align*} N \geq \frac{2^{2n}}{2n}. \end{align*} Combining, \begin{align*} \frac{2^{2n}}{2n} \leq N \leq 2^{2n}, \end{align*} proving (i). [/step]

[step:Prove (ii) by counting $p$ in the numerator and denominator of $N = \frac{(2n)!}{(n!)^2}$]Fix a prime $p$ with $n < p \leq 2n$. Write \begin{align*} N &= \binom{2n}{n} = \frac{(2n)!}{n! \, n!}. \end{align*} We compute $\nu_p((2n)!)$ and $\nu_p(n!)$ directly from the factorials. For the numerator: among the integers $1, 2, \ldots, 2n$, the multiples of $p$ are exactly $p, 2p, 3p, \ldots, \lfloor 2n/p \rfloor \cdot p$. Since $n < p \leq 2n$, we have $1 \leq 2n/p < 2$, so $\lfloor 2n/p \rfloor = 1$, meaning $p$ is the unique multiple of $p$ in $\{1, \ldots, 2n\}$. Since $p^2 > 2n$ (because $p > n \geq 1$ gives $p^2 > n \cdot p > n \cdot n \geq n$ — more directly, $p^2 > p \cdot n \geq 2n$ when $p \geq 2$, which holds), no element of $\{1, \ldots, 2n\}$ is divisible by $p^2$. Therefore \begin{align*} \nu_p((2n)!) = 1. \end{align*} For the denominator: among the integers $1, 2, \ldots, n$, none is divisible by $p$ since $p > n$. Therefore $\nu_p(n!) = 0$. Substituting, \begin{align*} \nu_p(N) = \nu_p((2n)!) - 2\nu_p(n!) = 1 - 0 = 1, \end{align*} proving (ii).[/step]

[guided]We count the exact power of $p$ dividing $N = (2n)! / (n!)^2$ by computing $\nu_p$ of each factorial separately; $\nu_p$ is additive on products and the formula \begin{align*} \nu_p(N) = \nu_p((2n)!) - 2\nu_p(n!) \end{align*} follows from $N \cdot (n!)^2 = (2n)!$. For $\nu_p((2n)!)$: we count how many of the integers $1, 2, \ldots, 2n$ are divisible by $p$, then by $p^2$, etc. The multiples of $p$ in $[1, 2n]$ are $p, 2p, \ldots$; the count is $\lfloor 2n/p \rfloor$. Since $p > n$, we have $2n/p < 2$; since $p \leq 2n$, we have $2n/p \geq 1$. Hence $\lfloor 2n/p \rfloor = 1$ and exactly one multiple of $p$ — namely $p$ itself — appears in $[1, 2n]$. Moreover $p^2 > p \cdot n \geq p \cdot (p/2) \cdot 2 = p^2$... more directly: since $p \geq n + 1$ and $p \geq 2$, we have $p^2 \geq 2p > 2n$, so no multiple of $p^2$ lies in $[1, 2n]$. Therefore $\nu_p((2n)!) = 1$. For $\nu_p(n!)$: since $p > n$, no integer in $[1, n]$ is divisible by $p$, so $\nu_p(n!) = 0$. Combining, $\nu_p(N) = 1 - 2 \cdot 0 = 1$. This says $p$ divides $N$ exactly once — a fact that will be crucial in the proof of Chebyshev's theorem, where the product $\prod_{n < p \leq 2n} p$ divides $N$ and is thereby bounded by $2^{2n}$.[/guided]

[step:Prove (iii) by showing $p$ appears with equal multiplicity in numerator and denominator] Fix a prime $p$ with $\frac{2n}{3} < p \leq n$. The hypothesis yields two bounds: \begin{align*} p \leq n \quad\text{and}\quad 2p \leq 2n < 3p, \end{align*} so $2n / p \in [2, 3)$, giving $\lfloor 2n / p \rfloor = 2$. Also $n/p \in [1, n/p]$ with $n/p \geq 1$ (since $p \leq n$) and $n/p < 3/2$ (since $p > 2n/3$ gives $n/p < 3/2$), so $\lfloor n/p \rfloor = 1$. The multiples of $p$ in $\{1, \ldots, 2n\}$ are $p$ and $2p$; the multiples of $p$ in $\{1, \ldots, n\}$ are just $p$. We now rule out higher powers of $p$. Since $p > 2n/3 \geq 2/3$, and $n \geq 1$, we have $p \geq 1$; more substantively, $p^2 > (2n/3)^2 = 4n^2 / 9$. For $n \geq 2$, this gives $p^2 > 4n^2/9 \geq 8n/9$; and since $p \geq 2$ whenever $p$ is prime, $p^2 \geq 2p > 4n/3 > n$, so $p^2 > n$. More directly: $p \geq 2$ and $p > 2n/3$ give $p^2 \geq 2p > 4n/3 \geq 2n$ provided $n \geq 3$. We need $p^2 > 2n$: from $p > 2n/3$ we have $p^2 > 2n \cdot p / 3 \geq 2n \cdot 2 / 3 = 4n/3$. For $p^2 > 2n$, write $p > 2n/3$ and use $p \geq 2$: then $p^2 \geq 2p > 4n/3$, which exceeds $2n$ iff $n < 0$ — so this estimate is insufficient on its own. We argue instead directly: if $p^2 \leq 2n$, then $p \leq \sqrt{2n}$, so $2n/3 < p \leq \sqrt{2n}$ forces $(2n/3)^2 < 2n$, i.e., $4n^2/9 < 2n$, i.e., $n < 9/2$. Thus for $n \geq 5$, we have $p^2 > 2n$; the remaining cases $n \in \{2, 3, 4\}$ are checked by hand (the range $(2n/3, n]$ is empty of primes for $n = 2$, contains only $p = 3$ for $n = 3$ with $p^2 = 9 > 6 = 2n$, and is empty of primes for $n = 4$ since $(8/3, 4] \cap \mathrm{primes} = \emptyset$). In all cases, $p^2 > 2n$, so $\lfloor 2n/p^2 \rfloor = \lfloor n/p^2 \rfloor = 0$. Therefore \begin{align*} \nu_p((2n)!) &= \sum_{i \geq 1} \left\lfloor \frac{2n}{p^i} \right\rfloor = 2 + 0 + 0 + \cdots = 2, \\ \nu_p(n!) &= \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor = 1 + 0 + 0 + \cdots = 1, \end{align*} where we used [Legendre's formula for prime-power valuation of factorials](/theorems/???). Substituting, \begin{align*} \nu_p(N) = \nu_p((2n)!) - 2\nu_p(n!) = 2 - 2 \cdot 1 = 0, \end{align*} so $p \nmid N$, i.e., $\gcd(p, N) = 1$, proving (iii). [/step]

[step:Prove (iv) via the identity $\lfloor 2x \rfloor - 2\lfloor x \rfloor \in \{0, 1\}$ and Legendre's formula]Fix a prime $p$ and an exponent $k \geq 1$ with $p^k \mid N$; we show $p^k \leq 2n$. By [Legendre's formula for the $p$-adic valuation of $m!$](/theorems/???), \begin{align*} \nu_p(m!) = \sum_{i \geq 1} \left\lfloor \frac{m}{p^i} \right\rfloor \end{align*} for every positive integer $m$ (the sum is finite: all terms with $p^i > m$ vanish). Applied to $m = 2n$ and $m = n$, \begin{align*} \nu_p(N) = \nu_p((2n)!) - 2 \nu_p(n!) = \sum_{i \geq 1} \left( \left\lfloor \frac{2n}{p^i} \right\rfloor - 2 \left\lfloor \frac{n}{p^i} \right\rfloor \right). \end{align*} We establish the elementary identity \begin{align*} \lfloor 2x \rfloor - 2 \lfloor x \rfloor \in \{0, 1\} \quad \text{for every } x \geq 0. \end{align*} [claim:For every $x \geq 0$, the quantity $\lfloor 2x \rfloor - 2\lfloor x \rfloor$ equals $0$ or $1$] [proof] Write $x = m + \alpha$ with $m = \lfloor x \rfloor \in \mathbb{Z}_{\geq 0}$ and $\alpha = x - \lfloor x \rfloor \in [0, 1)$. Then $2x = 2m + 2\alpha$. Since $2m$ is an integer and $2\alpha \in [0, 2)$, $\lfloor 2x \rfloor = 2m + \lfloor 2\alpha \rfloor$. Therefore \begin{align*} \lfloor 2x \rfloor - 2\lfloor x \rfloor = (2m + \lfloor 2\alpha \rfloor) - 2m = \lfloor 2\alpha \rfloor \in \{0, 1\}, \end{align*} since $2\alpha \in [0, 2)$ so $\lfloor 2\alpha \rfloor \in \{0, 1\}$. [/proof] [/claim] Applying the claim with $x = n / p^i$ for each $i \geq 1$ gives $\lfloor 2n/p^i \rfloor - 2 \lfloor n/p^i \rfloor \in \{0, 1\}$. Moreover, when $p^i > 2n$, we have $2n/p^i < 1$, so $\lfloor 2n/p^i \rfloor = 0$ and $\lfloor n/p^i \rfloor = 0$, contributing $0$ to the sum. Therefore \begin{align*} \nu_p(N) = \sum_{i \geq 1} \left( \left\lfloor \frac{2n}{p^i} \right\rfloor - 2 \left\lfloor \frac{n}{p^i} \right\rfloor \right) \leq \sum_{\substack{i \geq 1 \\ p^i \leq 2n}} 1 = \# \{i \geq 1 : p^i \leq 2n\}. \end{align*} The largest $i$ with $p^i \leq 2n$ is $i = \lfloor \log(2n) / \log p \rfloor$, so the count equals $\lfloor \log(2n)/\log p \rfloor$. Hence \begin{align*} \nu_p(N) \leq \left\lfloor \frac{\log(2n)}{\log p} \right\rfloor. \end{align*} Now suppose $p^k \mid N$, i.e., $\nu_p(N) \geq k$. Then \begin{align*} k \leq \nu_p(N) \leq \left\lfloor \frac{\log(2n)}{\log p} \right\rfloor \leq \frac{\log(2n)}{\log p}, \end{align*} so $k \log p \leq \log(2n)$, i.e., $p^k \leq 2n$, proving (iv).[/step]

[guided]The goal is to bound the prime-power factor: if $p^k$ divides $N = \binom{2n}{n}$, we want $p^k \leq 2n$. Legendre's prime-power valuation formula \begin{align*} \nu_p(m!) = \sum_{i \geq 1} \left\lfloor \frac{m}{p^i} \right\rfloor \end{align*} expresses the $p$-adic valuation of $m!$ as a sum of floor terms (the $i$-th counts multiples of $p^i$ in $\{1, \ldots, m\}$). Applied to the factorials in $N = (2n)! / (n!)^2$, we obtain \begin{align*} \nu_p(N) = \sum_{i \geq 1} \left( \left\lfloor \frac{2n}{p^i} \right\rfloor - 2 \left\lfloor \frac{n}{p^i} \right\rfloor \right). \end{align*} The key observation is that each summand is either $0$ or $1$. We prove the general fact: for any real $x \geq 0$, the value $\lfloor 2x \rfloor - 2\lfloor x \rfloor$ is either $0$ or $1$. Writing $x = m + \alpha$ with $m = \lfloor x \rfloor$ and $\alpha \in [0, 1)$ the fractional part: we have $2x = 2m + 2\alpha$ with $2\alpha \in [0, 2)$, so $\lfloor 2x \rfloor = 2m + \lfloor 2\alpha \rfloor$ and \begin{align*} \lfloor 2x \rfloor - 2 \lfloor x \rfloor = \lfloor 2\alpha \rfloor, \end{align*} which is $0$ if $\alpha < 1/2$ and $1$ if $\alpha \geq 1/2$. So the identity holds. Applied term by term, \begin{align*} \nu_p(N) \leq \#\{i \geq 1 : \text{term at index } i \text{ is nonzero}\} \leq \#\{i \geq 1 : p^i \leq 2n\}, \end{align*} where the last step uses that once $p^i > 2n$, both $\lfloor 2n/p^i \rfloor$ and $\lfloor n/p^i \rfloor$ are $0$, making the summand vanish. The count $\#\{i \geq 1 : p^i \leq 2n\}$ equals the largest $i_0$ with $p^{i_0} \leq 2n$, namely $i_0 = \lfloor \log(2n)/\log p \rfloor$. So $\nu_p(N) \leq i_0$. If $p^k \mid N$, then $k \leq \nu_p(N) \leq i_0$, hence $p^k \leq p^{i_0} \leq 2n$. This is exactly (iv). The intuition: the $p$-adic valuation of $N$ counts contributions from the $i$-th level $p^i$, each contributing at most $1$ and vanishing once $p^i > 2n$. So the full valuation is bounded by the number of levels still "active," which is the largest $i$ with $p^i \leq 2n$.[/guided]