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  "body": "\n## Statement\n\nLet $1/\\Gamma$ be the [[reciprocal gamma function|PY6K9AWR33]] and let $\\gamma$ be the [[Euler–Mascheroni constant|00C5P9JZBK]].\nFor every $z \\in \\C$,\n$$\\frac{1}{\\Gamma(z)} = z \\, e^{\\gamma z} \\prod_{n=1}^{\\infty} \\left(1 + \\frac{z}{n}\\right) e^{-z/n}.$$\nThe [[infinite product|RZ0Z2BTMT4]] converges absolutely and uniformly on every compact subset of $\\C$.\n\n## Remarks\n\nEach factor is $1 + O(z^2/n^2)$, and $\\sum 1/n^2 \u003c \\infty$ gives the convergence.\nThe partial product up to $n$ equals $\\frac{z(z+1)\\cdots(z+n)}{n!}\\, e^{z(\\gamma - H_n)}$, where $H_n$ is the [[harmonic number|0NANDX7V02]].\nSince $H_n - \\ln n \\to \\gamma$, this is the reciprocal of the quotient in the [[Euler–Gauss limit formula|SKRP9NDXR4]] times a factor tending to $1$, which proves the identity.\nThe product exhibits the simple zeros of $1/\\Gamma$ at $0, -1, -2, \\ldots$; it is the [[Weierstrass factorization|11VC51YHQJ]] of this entire function of order $1$.\nMultiplying the products for $1/\\Gamma(z)$ and $1/\\Gamma(-z)$ and using [[Euler's sine product|DQEHD56649]] gives $\\Gamma(z)\\Gamma(-z) = -\\pi/(z\\sin \\pi z)$, a form of the reflection formula.\n\n## Explanations\n\n### L1 human\n\nThe reciprocal of the gamma function can be written as an infinite product, much like a polynomial is a product of factors for its roots.\nHere the roots are $0, -1, -2, \\ldots$, and the exponential factors are there to make the infinite product converge.\n\n### L2 human\n\nTaking the logarithmic derivative of the product gives the series $\\psi(z) = -\\gamma - \\frac1z + \\sum_{n \\ge 1}\\left(\\frac1n - \\frac{1}{z+n}\\right)$ for the [[digamma function|HWRRXB0YXS]].\nDifferentiating once more gives $(\\ln\\Gamma)''(x) = \\sum_{n \\ge 0} (x+n)^{-2} \u003e 0$ on $(0, \\infty)$, so $\\ln \\Gamma$ is convex there.\nThe product also yields the Taylor expansion of $\\ln\\Gamma(1 + z)$ around $0$, whose coefficients are values of the [[Riemann zeta function|1XWBY4N0VT]] at the integers $k \\ge 2$.\n",
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