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  "body": "\n## Proof\n\nWe integrate by parts on $[c, d] \\subset (0, \\infty)$ and let $c \\to 0^+$ and $d \\to \\infty$. Then we argue by induction on $n$.\n\n1. Setup. For $t \u003e 0$ and real $r$, $t^r = e^{r \\ln t}$ is the [[real power|0TV07AQT62]], where $e^y = \\exp(y)$ is the [[exponential function|0RMVS7RXZX]] and $\\ln$ is the [[natural logarithm|0044CPHJBV]].\n2. For $s \u003e 0$, the integral of the [[gamma function|0V3FZMVB4K]] is the sum $\\Gamma(s) = A_s + B_s$ of two [[improper integrals|0VCTBJMEJF]]. Here $A_s = \\lim_{c \\to 0^+} \\int_c^1 t^{s-1} e^{-t} \\, dt$ and $B_s = \\lim_{d \\to \\infty} \\int_1^d t^{s-1} e^{-t} \\, dt$. The definition of $\\Gamma(s)$ presupposes that both limits exist.\n3. Facts on $\\exp$. It is differentiable with $\\exp' = \\exp$ ([[derivative|0EX9GFJ0B3]]). It satisfies $e^{x+y} = e^x e^y$ ([[sum rule|0E8HNVBCSP]]), so $e^{-b} = 1/e^b$. It is [[positive|0P9BNBW2W3]] and [[strictly increasing|0KN2V7840E]].\n4. By the [[chain rule|0RXX2PT1MQ]] and step 3, $t \\mapsto -e^{-t}$ has derivative $e^{-t}$. By the [[derivative of $x^a$|0KY08TWDMQ]], $t \\mapsto t^s$ has derivative $s t^{s-1}$ on $(0, \\infty)$.\n5. All the functions in step 4 and their derivatives are differentiable on $(0, \\infty)$, hence [[continuous|07ZMX4ANJS]]. So their [[products are continuous|0VGMHCF4SC]] and [[Riemann integrable|06SESV7GQY]] on each $[c, d] \\subset (0, \\infty)$.\n6. Let $s \u003e 0$ and $0 \u003c c \u003c d$. Apply [[integration by parts|08Y1BBQTK9]] on $[c, d]$ with $u(t) = t^s$ and $v(t) = -e^{-t}$. Steps 4 and 5 give\n$$\\int_c^d t^s e^{-t} \\, dt = c^s e^{-c} - d^s e^{-d} + s \\int_c^d t^{s-1} e^{-t} \\, dt.$$\n7. Lemma A: $c^s \\to 0$ as $c \\to 0^+$. Let $\\eta \u003e 0$ and $\\delta = e^{(\\ln \\eta)/s} \u003e 0$. By the [[inverse relations|0RKW2X63YR]], $\\ln \\delta = (\\ln \\eta)/s$.\n8. For $0 \u003c c \u003c \\delta$, $\\ln c \u003c \\ln \\delta$, since $\\ln$ is strictly increasing by the same result. So $s \\ln c \u003c \\ln \\eta$, and step 3 gives $0 \u003c c^s = e^{s \\ln c} \u003c e^{\\ln \\eta} = \\eta$. This proves Lemma A.\n9. Lemma B: $d^s e^{-d} \\to 0$ as $d \\to \\infty$. By the [[Archimedean property|0A74SX55QP]], choose a natural number $n \\ge s$.\n10. For $d \\ge 1$, $\\ln d \\ge \\ln 1 = 0$, so $s \\ln d \\le n \\ln d$. By steps 3 and 1, $d^s \\le e^{n \\ln d}$. By induction on $n$ with the sum rule, $e^{n \\ln d} = (e^{\\ln d})^n$. This equals $d^n$ by the inverse relations.\n11. So $0 \u003c d^s e^{-d} \\le d^n / e^d$ for $d \\ge 1$. Let $\\eta \u003e 0$. Since [[$e^d / d^n \\to \\infty$|0CT6CBGTYY]], there is $X \\ge 1$ with $e^d / d^n \u003e 1/\\eta$ for $d \u003e X$. Then $d^s e^{-d} \u003c \\eta$ for $d \u003e X$. This proves Lemma B.\n12. Take $(c, d) = (c, 1)$ in step 6, with $0 \u003c c \u003c 1$. Since $1^s = e^{s \\ln 1} = 1$, we get $\\int_c^1 t^s e^{-t} \\, dt = c^s e^{-c} - e^{-1} + s \\int_c^1 t^{s-1} e^{-t} \\, dt$.\n13. As $c \\to 0^+$, $e^{-c} \\to e^0 = 1$ by continuity, and $c^s \\to 0$ by Lemma A. The right side of step 12 then tends to $-e^{-1} + s A_s$. This uses step 2 and the [[algebra of limits|07XBM3A9J5]]. So $A_{s+1}$ exists and $A_{s+1} = -e^{-1} + s A_s$.\n14. Take $(c, d) = (1, d)$ in step 6, with $d \u003e 1$. We get $\\int_1^d t^s e^{-t} \\, dt = e^{-1} - d^s e^{-d} + s \\int_1^d t^{s-1} e^{-t} \\, dt$.\n15. As $d \\to \\infty$, Lemma B, step 2 and the algebra of limits give $B_{s+1} = e^{-1} + s B_s$.\n16. Adding steps 13 and 15 gives $\\Gamma(s + 1) = A_{s+1} + B_{s+1} = s (A_s + B_s) = s \\, \\Gamma(s)$.\n17. The value $\\Gamma(1)$. Here $t^0 = e^0 = 1$, so the integrand is $e^{-t}$. Its antiderivative is $-e^{-t}$ by step 4. The [[second fundamental theorem of calculus|0XYAFCN0PT]] gives $\\int_c^1 e^{-t} \\, dt = e^{-c} - e^{-1}$ and $\\int_1^d e^{-t} \\, dt = e^{-1} - e^{-d}$.\n18. As $c \\to 0^+$, $e^{-c} \\to 1$, so $A_1 = 1 - e^{-1}$. By Lemma B with $s = 1$ and $n = 1$, $d e^{-d} \\to 0$. For $d \\ge 1$, $0 \u003c e^{-d} \\le d e^{-d}$, so $e^{-d} \\to 0$ and $B_1 = e^{-1}$.\n19. So $\\Gamma(1) = A_1 + B_1 = 1$.\n20. We prove $\\Gamma(n + 1) = n!$ by induction on the natural number $n$. For $n = 0$, step 19 gives $\\Gamma(1) = 1 = 0!$, by the definition of the [[factorial|0WFCGVMRRC]].\n21. Assume $\\Gamma(n + 1) = n!$. Step 16 with $s = n + 1 \u003e 0$ gives $\\Gamma(n + 2) = (n + 1) \\Gamma(n + 1) = (n + 1) \\cdot n! = (n + 1)!$.\n22. By induction, $\\Gamma(n + 1) = n!$ for every natural number $n$.\n\n## Idea\n\nIntegration by parts moves one factor of $t$ from $t^s$ onto the derivative, which gives $s t^{s-1}$. The boundary term $t^s e^{-t}$ vanishes at both ends: $t^s \\to 0$ at $0$ because $s \u003e 0$, and $e^{-t}$ beats any power at infinity. Together with $\\Gamma(1) = 1$, the recursion $\\Gamma(s+1) = s \\Gamma(s)$ gives the factorials.\n",
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