The product of all natural numbers

Note that for the zeta function $\zeta(s)=\sum_{n=1}^{\infty}n^{-s},$ we have that $\zeta'(s)=-\sum_{n=1}^{\infty}n^{-s}\ln n,$ so that $\zeta'(0)=-\sum_{n=1}^{\infty}\ln n=-\frac{1}{2}\ln 2\pi$ after using a known value of the zeta function derivative (gotten from the Hadamard product or Wallis formula). Then we may evaluate $1\times 2\times 3\times\cdots=\exp(-\zeta'(0))=(2\pi)^{1/2}.$

EDIT: More information on the Wallis formula $\frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{(2n)^2}{(2n-1)(2n+1)}=\frac{2\cdot 2}{1\cdot 3}\frac{4\cdot 4}{3\cdot 5}\frac{6\cdot 6}{5\cdot 7}\cdots.$ This product emerges when we examine the function $F(s)=(1-2^{1-s})\zeta(s)=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right).$ Taking the $s$ derivative of both sides, we have $2^{1-s}(\ln 2)\zeta(s)+(1-2^{1-s})\zeta'(s)=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^n\left[\frac{\ln n}{n^s}-\frac{\ln(n+1)}{(n+1)^s}\right].$ Evaluating at $s=0$ and using $\zeta(0)=-1/2$ , we have $2(\ln 2)(-1/2)+(1-2)\zeta'(0)=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^n\left[\ln n-\ln(n+1)\right]=\frac{1}{2}\ln\left(\frac{2\cdot 2}{1\cdot 3}\frac{4\cdot 4}{3\cdot 5}\frac{6\cdot 6}{5\cdot 7}\cdots\right)=\frac{1}{2}\ln\frac{\pi}{2}.$ From this we see that $\zeta'(0)=-\tfrac{1}{2}\ln 2\pi$ .