Let us consider the set $[n] = \{1, 2, \dots, n\}.$ The binomial identity $\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$ can be interpreted combinatorially by summing $(-1)^{|S|}$ over all subsets $S \subseteq X$. In other words, we assign a sign of $+1$ to each subset of even size and $-1$ to each subset of odd size, then sum all these signs. The goal is to show that the total signed count is zero. To prove this, we define a sign-reversing involution on the power set $\mathcal{P}(X)$. For each subset $S \subseteq X$, we define a map $\varphi(S)$ that toggles the presence of the element $n$. That is, if $n \in S$, then $\varphi(S) = S \setminus \{n\}$; otherwise, $\varphi(S) = S \cup \{n\}$. This map is an involution because applying it twice returns the original set: $\varphi(\varphi(S)) = S$. It also has no fixed points, since every set either gains or loses the element $n$, so $\varphi(S) \neq S$ for all $S$. Crucially, the involution pairs each subset $S$ with another subset $\varphi(S)$ that differs in size by one. This means the signs $(-1)^{|S|}$ and $(-1)^{|\varphi(S)|}$ are opposite, and so each such pair contributes zero to the total sum. Since every subset of $X$ is in exactly one such pair, the entire sum cancels out. This completes the proof.