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  "body": "\n## Proof\n\nWe argue by induction on $n$.\n\n1. Fix a real number $x \u003e 0$. For a natural number $n$, the power $x^n$ is the product of $n$ factors equal to $x$, with $x^0 = 1$. So $x^n \u003e 0$, and $\\ln(x^n)$ is defined by the [[definition of the natural logarithm|0044CPHJBV]].\n2. Base case $n = 0$. By the [[definition of the exponential function|0RMVS7RXZX]], $e^0 = 1$, since every term of the series after the first vanishes at $0$.\n3. Since [[$\\ln(e^y) = y$ for every real $y$|0RKW2X63YR]], we get $\\ln(x^0) = \\ln 1 = \\ln(e^0) = 0 = 0 \\cdot \\ln x$.\n4. Induction step. Assume that $\\ln(x^n) = n \\ln x$ for some natural number $n$.\n5. We have $x^{n+1} = x^n \\cdot x$, and both factors are positive.\n6. By the [[logarithm of a product|0K2FNY6DYV]], $\\ln(x^{n+1}) = \\ln(x^n) + \\ln x$.\n7. By the induction hypothesis, this equals $n \\ln x + \\ln x = (n+1) \\ln x$.\n8. By induction, $\\ln(x^n) = n \\ln x$ for every natural number $n$ and every real $x \u003e 0$.\n\n## Idea\n\nA power $x^n$ is a product of $n$ copies of $x$. The logarithm turns each product into a sum, so it turns $n$ copies of $x$ into $n$ copies of $\\ln x$. Induction makes this count precise.\n",
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