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  "body": "\n## Proof\n\nWe argue by induction on $n \\ge 1$. Write $p_n(x) = x^n$, and use the convention $x^0 = 1$.\n\n1. Base case $n = 1$. Fix $x \\in \\R$. For $h \\ne 0$, the difference quotient is $\\frac{(x + h) - x}{h} = 1$.\n2. A function that is constantly $1$ near $0$ has limit $1$ at $0$. Indeed, in the [[epsilon-delta form of limits|0GBZDAFPX8]], any $\\delta$ works since $\\abs{1 - 1} = 0 \u003c \\varepsilon$.\n3. So $p_1$ has [[derivative|0F1WR9C83K]] $1 = 1 \\cdot x^0$ at $x$. Thus the claim holds for $n = 1$.\n4. Induction step. Assume that $p_n$ is [[differentiable|0A61B60B5R]] at every $x$ with $p_n'(x) = n x^{n-1}$, for some $n \\ge 1$.\n5. For all $x$, $p_{n+1}(x) = x^n \\cdot x = p_n(x) \\, p_1(x)$.\n6. By step 4, step 3 and the [[product rule|0V9A43H7WV]], $p_{n+1}$ is differentiable at every $x$, and\n$$p_{n+1}'(x) = p_n'(x) \\, p_1(x) + p_n(x) \\, p_1'(x) = n x^{n-1} \\cdot x + x^n \\cdot 1.$$\n7. The right side equals $n x^n + x^n = (n + 1) x^n$. This is the claim for $n + 1$.\n8. By induction, $x \\mapsto x^n$ is differentiable on $\\R$ with derivative $n x^{n-1}$ for every $n \\ge 1$.\n\n## Idea\n\nThe case $n = 1$ is a direct computation. Writing $x^{n+1} = x^n \\cdot x$, the product rule turns the formula for $n$ into the formula for $n + 1$.\n",
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