Factorial of infinity So, I've read in this article that:
$$\zeta'(0) = \log\sqrt\frac 1 {2\pi}$$
And that:
$$e^{-\zeta'(0)} = 1\cdot2\cdot3\cdots \infty = \infty! = \sqrt{2\pi}$$
I found this result very strange and tried to simplify if but i couldn't. Can somebody help me?
Thanks!

One way to crosscheck (not prove/disprove (!)) that results approximative and numerically is to use the alternating zeta at the same parameter, which can be summed using simple summation procedures for divergent summation. If we write
$$ \eta(s) = \frac 1{1^s} - \frac 1{2^s} + \frac 1{3^s} - \cdots $$
and furtherly 
$$ \eta(s) = \exp(-s\ln 1) - \exp(- s\ln2) + \exp(-s\ln 3) - \cdots $$
then we can write the derivative as
$$ \eta(s)' = (-\ln1)\cdot \exp(-s\ln 1) - (-\ln2)\cdot \exp(- s\ln2) + (-\ln3)\cdot \exp(-s\ln 3) - \cdots $$
and letting $\lim s \to 0$ this gives
$$ \begin{eqnarray} 
\eta(0)' &=& (-\ln1) - (-\ln2) + (-\ln3) - \cdots  \\
 \eta(0)' &=& - (\ln1 - \ln2 + \ln3 - \cdots) \tag 1 \end{eqnarray}$$
The latter can be handled by a simple procedure for divergent summation; for instance Euler-summation $ \mathfrak E $ which gives then
$$ \eta(0)'  \underset{\mathfrak E}{=} 0.225791352645\ldots $$
Now in (1) we see the alternating signs but for the equivalent value of the zeta-function this series is non-alternating. So
$$ \zeta(0)' = - (\ln1 + \ln2 + \ln3 + \cdots) \tag 2$$
We assume now, that the value $ \zeta(0)' = s$ exists and that we can handle it into the divergent sum. Here we allude to Euler's trick of the transformation-formula between zeta() and eta() and exploit that the logs at even arguments are negative in the eta-formula. But $ \ln(2m) = \ln2 + \ln m$ and thus we might write
$$ \begin{eqnarray} 
 s &=& - ( &(\ln 1 + \ln 2 + \ln 3 + \ln 4 + \ln 5 + \ln 6 + \ln 7\cdots) \\
 s &=& - ( &(\ln 1 - \ln 2 + \ln 3 - \ln 4 + \ln 5 - \ln 6 + \ln 7\cdots) \\
         & & &+2 \cdot ( \ln 2 + \ln 4 + \ln 6 + \ln 8 + \cdots) ) \\
 s&=& &\eta(0)' - 2 \cdot ( (\ln2 + \ln1) + (\ln2 + \ln 2) + (\ln 2 + \ln 3) + \cdots)) \\
 s&=& &\eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \tag 3\\
\end{eqnarray} $$
where we have accumulated the constant $\ln 2$ which occurs as summand at each consecutive logarithm in the last equation (3) as $\zeta(0) \cdot \ln 2$. Then we have a determination for $s$ by
$$ \begin{eqnarray} &s&=& \eta(0)' - 2 \cdot ( \zeta(0) \cdot \ln 2 - s ) \\
 &-s&=& \eta(0)' - 2 \cdot \zeta(0) \cdot \ln 2  \\
  \zeta(0)'&=s&=& -\eta(0)' - \ln 2  \tag 4 \end{eqnarray}$$
$$\begin{eqnarray} &\zeta(0)' &\sim - 0.225791352645\ldots - 0.693147180560 \\ 
 && \sim -0.918938533205\ldots \tag 5 \\
 \end{eqnarray}$$
And indeed, numerically we get using Pari/GP
log(1/sqrt(2*Pi))
 %393 = -0.918938533205

and also
h=1e-12;(zeta(0+h/2)-zeta(0-h/2))/h
 %394 = -0.918938533205

This all required assumptions about the general summability of the divergent terms. But there is a fairly well developed theory of divergent summation, from which one may take the confirmation about the acceptability of the above transformations. They are intended to show one path to obtain the result; perhaps/likely there are others more elegant or even rigid ones.        

Appendix: The Pari/GP code for the numerical evaluation of $\eta(0)'$ in (3) is
-sumalt(k=0,(-1)^k*log(1+k))
 %387 = 0.225791352645

