Weierstrass product for the gamma function

Remarks

Each factor is $1 + O(z^2/n^2)$, and $\sum 1/n^2 < \infty$ gives the convergence. The 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. Since $H_n - \ln n \to \gamma$, this is the reciprocal of the quotient in the Euler–Gauss limit formula times a factor tending to $1$, which proves the identity. The product exhibits the simple zeros of $1/\Gamma$ at $0, -1, -2, \ldots$; it is the Weierstrass factorization of this entire function of order $1$. Multiplying the products for $1/\Gamma(z)$ and $1/\Gamma(-z)$ and using Euler's sine product gives $\Gamma(z)\Gamma(-z) = -\pi/(z\sin \pi z)$, a form of the reflection formula.

Going deeper

For a student

Taking 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. Differentiating once more gives $(\ln\Gamma)''(x) = \sum_{n \ge 0} (x+n)^{-2} > 0$ on $(0, \infty)$, so $\ln \Gamma$ is convex there. The product also yields the Taylor expansion of $\ln\Gamma(1 + z)$ around $0$, whose coefficients are values of the Riemann zeta function at the integers $k \ge 2$.