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  "body": "\n## Proof\n\nWe bound $e^x$ below by a single term of its series.\n\n1. Let $n$ be a natural number. Here $\\lim_{x \\to \\infty} \\frac{e^x}{x^n} = \\infty$ means the following. For each real $M$ there is $X$ with $\\frac{e^x}{x^n} \u003e M$ for all $x \u003e X$.\n2. Let $x \u003e 0$. By the definition of the [[exponential function|0RMVS7RXZX]], $e^x = \\sum_{k=0}^{\\infty} \\frac{x^k}{k!}$. The series converges, since its [[radius of convergence is infinite|0PVJRACB9H]].\n3. Every term $\\frac{x^k}{k!}$ is nonnegative. By the [[result on series of nonnegative terms|0MX1HZBBXJ]], the sum is the supremum of the partial sums. So $e^x \\ge \\sum_{k=0}^{n+1} \\frac{x^k}{k!}$.\n4. The terms with $k \\le n$ in this partial sum are nonnegative. So $e^x \\ge \\frac{x^{n+1}}{(n+1)!}$.\n5. Divide by $x^n \u003e 0$. This gives $\\frac{e^x}{x^n} \\ge \\frac{x}{(n+1)!}$ for all $x \u003e 0$.\n6. Let $M$ be real, and let $X = \\max(0, M \\cdot (n+1)!)$. For $x \u003e X$, we have $x \u003e 0$ and $\\frac{x}{(n+1)!} \u003e M$.\n7. By step 5, $\\frac{e^x}{x^n} \u003e M$ for all $x \u003e X$. So $\\frac{e^x}{x^n} \\to \\infty$ as $x \\to \\infty$.\n\n## Idea\n\nFor positive $x$, all terms of the exponential series are positive. Keeping only the term $x^{n+1}/(n+1)!$ already beats $x^n$ by a factor that grows like $x$.\n",
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