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  "body": "\n## Proof\n\nWe study the sign of the derivative of $g(x) = e^x - 1 - x$ on each side of $0$.\n\n1. Let $g(x) = e^x - 1 - x$ for real $x$, where $e^x = \\exp(x)$ is the [[exponential function|0RMVS7RXZX]]. The claim is that $g(x) \\ge 0$, with equality only at $x = 0$.\n2. From the series, $e^0 = 1$, so $g(0) = 0$.\n3. By [[the derivative of exp|0EX9GFJ0B3]], the [[derivative of a polynomial|0MH1WK89PP]] and the [[linearity of differentiation|0M0TKFRX2A]], $g$ is differentiable on $\\R$ and $g'(x) = e^x - 1$.\n4. So $g$ is [[continuous|07ZMX4ANJS]] on $\\R$.\n5. Since $\\exp$ is [[strictly increasing|0KN2V7840E]] and $e^0 = 1$, we have $g'(x) \u003e 0$ for $x \u003e 0$ and $g'(x) \u003c 0$ for $x \u003c 0$.\n6. On $I = [0, \\infty)$, the interior points are the $x \u003e 0$, where $g'(x) \u003e 0$. By the theorem on a [[positive derivative|07J5X2T1ZX]], $g$ is strictly increasing on $[0, \\infty)$.\n7. So for $x \u003e 0$, $g(x) \u003e g(0) = 0$.\n8. Let $h = -g$. By linearity of differentiation, $h'(x) = 1 - e^x$, which is positive for $x \u003c 0$ by step 5. Also $h$ is continuous by step 4.\n9. On $I = (-\\infty, 0]$, the interior points are the $x \u003c 0$. By the theorem on a positive derivative, $h$ is strictly increasing on $(-\\infty, 0]$.\n10. So for $x \u003c 0$, $h(x) \u003c h(0) = 0$, that is, $g(x) \u003e 0$.\n11. By steps 2, 7 and 10, $g(x) \\ge 0$ for all real $x$, with $g(x) = 0$ only for $x = 0$. So $1 + x \\le e^x$, with equality if and only if $x = 0$.\n\n## Idea\n\nThe function $e^x - 1 - x$ vanishes at $0$. Its derivative $e^x - 1$ is negative to the left of $0$ and positive to the right. So the function falls until $0$ and rises after it, and its only zero is at $0$.\n",
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