How is it possible that $\infty!=\sqrt{2\pi}$? I read from here that:
$$\infty!=\sqrt{2\pi}$$
How is this possible ?
$$\infty!=1\times2\times3\times4\times5\times\ldots$$
But
\begin{align}
1&=1\\
1\times2&=2\\
1\times2\times3&=6\\
&~\vdots\\
1\times2\times3\times\ldots\times50&=3.0414093201713376\times10^{64}
\end{align}
This is obviously increasing, so how does factorial of $\infty$ become $\sqrt{2\pi}$ ?

It is taken from 
$$ 1\cdot2\cdot3\cdot \ldots \cdot n= n!$$
This is the exponential of
$$ \ln(1)+\ln(2)+\ln(3)+ \ldots + \ln(n) = \ln(n!) $$
Now if you write formally the derivative of the Dirichlet-series for zeta then you have
$$ \zeta'(s) = {\ln(1) \over 1^s}+{\ln(1/2) \over 2^s}  +{\ln(1/3) \over 3^s} + \ldots $$
This is for some s convergent and from there can be analytically continued to $s=0$ as well from where the the formal expression reduces to
$$ \zeta'(0) = -(\ln(1) +\ln(2) +\ln(3)  + \ldots )$$
which is then formally identical to $ - \lim_{n \to \infty} \ln(n!)$ .   
Now the slope of zeta at zero can numerically be approximated and gives a concrete number $\zeta'(0) \approx -0.91893...$. It can also analytically be determined to equal $\ln(1/\sqrt{2\pi})$ .     
Finally, since the formal notations coincide (except of the sign) one goes to write the exponential of this value to become the "regularized" value of the infinite factorial.
