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  "body": "\n## Proof\n\n1. Let $n \\ge 1$. The integers $\\Z$ form a commutative ring. Apply the [[binomial theorem|0EHWPWGHB2]] in $\\Z$ with $x = -1$ and $y = 1$.\n2. It gives $(-1 + 1)^n = \\sum_{k=0}^{n} \\binom{n}{k} (-1)^k 1^{n-k} = \\sum_{k=0}^{n} (-1)^k \\binom{n}{k}$.\n3. The left side is $0^n$. Since $n \\ge 1$, we have $0^n = 0 \\cdot 0^{n-1} = 0$.\n4. Hence $\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} = 0$.\n\n## Idea\n\nPutting $x = -1$ and $y = 1$ in the binomial theorem gives each coefficient the sign $(-1)^k$.\nThe left side becomes $0^n$, which is $0$ as soon as $n \\ge 1$.\n",
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