[proofplan]
We count all subsets of a fixed $2n$-element set by their cardinality. This gives the identity that the sum of the binomial coefficients in row $2n$ is $4^n$. The upper bound follows because the central [binomial coefficient](/page/Binomial%20Coefficient) is one nonnegative term in that sum, while the lower bound follows after proving directly from consecutive ratios that the central coefficient is the largest term in the row.
[/proofplan]
[step:Count all subsets by cardinality]
Fix $n \in \mathbb{N}$, and define the finite set $A := \{1,\dots,2n\}$. For each integer $k$ with $0 \leq k \leq 2n$, define
\begin{align*}
\mathcal{C}_k := \{E \subset A : |E| = k\}.
\end{align*}
By the definition of the binomial coefficient, $|\mathcal{C}_k| = \binom{2n}{k}$. The families $\mathcal{C}_0,\dots,\mathcal{C}_{2n}$ are pairwise disjoint and their union is the power set $\mathcal{P}(A)$, because every subset of $A$ has exactly one cardinality between $0$ and $2n$. Therefore
\begin{align*}
|\mathcal{P}(A)| = \sum_{k=0}^{2n} |\mathcal{C}_k| = \sum_{k=0}^{2n} \binom{2n}{k}.
\end{align*}
Each element of $A$ has two independent choices, membership or non-membership in a subset, so $|\mathcal{P}(A)| = 2^{2n} = 4^n$. Hence
\begin{align*}
\sum_{k=0}^{2n} \binom{2n}{k} = 4^n.
\end{align*}
[/step]
[step:Use nonnegativity of the binomial sum to get the upper bound]
For every integer $k$ with $0 \leq k \leq 2n$, the number $\binom{2n}{k}$ is the cardinality of $\mathcal{C}_k$, so it is nonnegative. Since $\binom{2n}{n}$ is one term of the sum computed above,
\begin{align*}
\binom{2n}{n} \leq \sum_{k=0}^{2n} \binom{2n}{k} = 4^n.
\end{align*}
This proves the upper bound.
[/step]
[step:Show that the central coefficient is the largest coefficient in row $2n$]
For each integer $k$ with $0 \leq k < 2n$, define the consecutive ratio
\begin{align*}
r_k := \frac{\binom{2n}{k+1}}{\binom{2n}{k}}.
\end{align*}
Using the factorial formula for binomial coefficients,
\begin{align*}
r_k = \frac{(2n)!}{(k+1)!(2n-k-1)!} \cdot \frac{k!(2n-k)!}{(2n)!}.
\end{align*}
Cancelling the common positive factors gives
\begin{align*}
r_k = \frac{2n-k}{k+1}.
\end{align*}
If $0 \leq k \leq n-1$, then $2n-k \geq k+1$, so $r_k \geq 1$. Thus
\begin{align*}
\binom{2n}{0} \leq \binom{2n}{1} \leq \cdots \leq \binom{2n}{n}.
\end{align*}
If $n \leq k < 2n$, then $2n-k \leq k+1$, so $r_k \leq 1$. Thus
\begin{align*}
\binom{2n}{n} \geq \binom{2n}{n+1} \geq \cdots \geq \binom{2n}{2n}.
\end{align*}
Combining the two chains, for every integer $k$ with $0 \leq k \leq 2n$,
\begin{align*}
\binom{2n}{k} \leq \binom{2n}{n}.
\end{align*}
[guided]
We need a reason that the middle term dominates every other term in the row. Rather than cite unimodality, we prove it from the ratio of consecutive binomial coefficients.
For each integer $k$ with $0 \leq k < 2n$, define
\begin{align*}
r_k := \frac{\binom{2n}{k+1}}{\binom{2n}{k}}.
\end{align*}
Using the factorial formula for binomial coefficients, we compute
\begin{align*}
r_k = \frac{(2n)!}{(k+1)!(2n-k-1)!} \cdot \frac{k!(2n-k)!}{(2n)!}.
\end{align*}
All factors are positive, and cancellation gives
\begin{align*}
r_k = \frac{2n-k}{k+1}.
\end{align*}
This ratio tells us whether the row is increasing or decreasing at position $k$. If $0 \leq k \leq n-1$, then $2n-k \geq k+1$, so $r_k \geq 1$. Therefore
\begin{align*}
\binom{2n}{k+1} \geq \binom{2n}{k}
\end{align*}
for every $k$ with $0 \leq k \leq n-1$. Hence the coefficients increase up to the central coefficient:
\begin{align*}
\binom{2n}{0} \leq \binom{2n}{1} \leq \cdots \leq \binom{2n}{n}.
\end{align*}
If $n \leq k < 2n$, then $2n-k \leq k+1$, so $r_k \leq 1$. Therefore
\begin{align*}
\binom{2n}{k+1} \leq \binom{2n}{k}
\end{align*}
for every $k$ with $n \leq k < 2n$. Hence the coefficients decrease after the central coefficient:
\begin{align*}
\binom{2n}{n} \geq \binom{2n}{n+1} \geq \cdots \geq \binom{2n}{2n}.
\end{align*}
The two monotonicity chains together show that the largest term in row $2n$ is the central term. Thus, for every integer $k$ with $0 \leq k \leq 2n$,
\begin{align*}
\binom{2n}{k} \leq \binom{2n}{n}.
\end{align*}
[/guided]
[/step]
[step:Compare the binomial sum with the largest term to get the lower bound]
There are exactly $2n+1$ integers $k$ satisfying $0 \leq k \leq 2n$. Since every term in the sum is at most the central term,
\begin{align*}
4^n = \sum_{k=0}^{2n} \binom{2n}{k} \leq \sum_{k=0}^{2n} \binom{2n}{n}.
\end{align*}
The right-hand side is a sum of $2n+1$ identical terms, so
\begin{align*}
\sum_{k=0}^{2n} \binom{2n}{n} = (2n+1)\binom{2n}{n}.
\end{align*}
Therefore
\begin{align*}
4^n \leq (2n+1)\binom{2n}{n}.
\end{align*}
Since $2n+1 > 0$, division by $2n+1$ gives
\begin{align*}
\frac{4^n}{2n+1} \leq \binom{2n}{n}.
\end{align*}
Together with the upper bound, this proves
\begin{align*}
\frac{4^n}{2n+1} \leq \binom{2n}{n} \leq 4^n.
\end{align*}
[/step]
