[proofplan] We evaluate the alternating binomial sum by viewing it as the binomial expansion of $(1 + x)^n$ at $x = -1$. The [binomial theorem](/theorems/750) identifies the finite polynomial expansion, and substitution gives the desired sum as $(1 - 1)^n$. The assumption $n \geq 1$ is exactly what ensures that this last quantity is $0$. [/proofplan] [step:Expand $(1+x)^n$ using the binomial theorem] Define the polynomial \begin{align*} p_n: \mathbb{R} &\to \mathbb{R}, \quad x \mapsto (1 + x)^n. \end{align*} By the Binomial Theorem (citing a result not yet in the wiki: Binomial Theorem), for every $x \in \mathbb{R}$, \begin{align*} p_n(x) = \sum_{k=0}^{n} \binom{n}{k} x^k. \end{align*} [guided] We want to recognize the given finite sum as the value of a polynomial. Define \begin{align*} p_n: \mathbb{R} &\to \mathbb{R}, \quad x \mapsto (1 + x)^n. \end{align*} The Binomial Theorem states that for every real number $x$ and every natural number $n$, the polynomial $(1+x)^n$ expands as \begin{align*} (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k. \end{align*} Here the hypothesis needed for the theorem is exactly that $n$ is a natural number, which is part of the statement. Therefore, for every $x \in \mathbb{R}$, \begin{align*} p_n(x) = \sum_{k=0}^{n} \binom{n}{k} x^k. \end{align*} This is useful because the factor $(-1)^k$ in the theorem statement appears by evaluating the expansion at $x=-1$. [/guided] [/step] [step:Evaluate the expansion at $x=-1$] Substituting $x=-1$ into the expansion gives \begin{align*} (1 + (-1))^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k. \end{align*} Since $1 + (-1) = 0$ and $n \geq 1$, we have $0^n = 0$. Hence \begin{align*} \sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0. \end{align*} This proves the claim. [/step]