Proof of $\sum\limits_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x$ as $x \to \infty$ Prove that
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x \,\,\,\text{as}\,\,\, x \to \infty$$
and
$$\sum_{n=1}^{\infty} \frac{(-x)^n \log(n!)}{n!} \to 0 \,\,\,\text{as}\,\,\, x \to \infty$$
This question is related to  my previous question. My heuristic approach is that the sum's major contribution comes from $n\approx x$ term, so
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim \frac{x^x\log(x!)}{x!}$$
but using the Stirling formula twice leads to 
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim \frac{\sqrt{x}\log(x)}{\sqrt{2\pi}}e^x$$
But this reasoning is flawed and I believe I should not ignore all the other terms. (Which I believe to account for $\sqrt{2\pi x}$factor)
What kind of approach may give the wanted asymptotic behavior? Any kind of hint is welcome.

Your idea is right that the entries near $n=x$ matter the most. In particular, the entries with $n=x+O(\sqrt{x})$ dominate this sum. 
The following is not a complete answer, but I believe the holes can be filled in. 
Suppose that $N=x + \alpha \sqrt{x},$ where $\alpha$ is fixed. Then, by Stirling's formula 
\begin{align*}
\frac{\ln(N!)}{N!} x^N &= \left(1+O\left(\frac{1}{\ln x}\right)\right) \frac{(x+\alpha \sqrt{x}) \ln x}{\sqrt{2 \pi x} (x+\alpha \sqrt{x})^{x+\alpha \sqrt{x}}}e^{x+\alpha \sqrt{x}} x^{x+\alpha \sqrt{x}} \\ &=\left(1+O\left(\frac{1}{\ln x}\right)\right) x (\ln x) e^x \left[e^{\alpha \sqrt{x}} \left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}} \frac{1}{\sqrt{2\pi x}} \right].
\end{align*}
Now, note that 
\begin{align*}
\left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}}&=\exp \left( -(x+\alpha \sqrt{x}) \cdot \ln\left(1+\frac{\alpha}{\sqrt{x}} \right)\right) \\&=\exp\left( -(x+\alpha \sqrt{x}) \left( \frac{\alpha}{\sqrt{x}} - \frac{\alpha^2}{2x} + O(n^{-3/2}) \right) \right) \\&= \exp \left( -\alpha \sqrt{x} - \alpha^2/2 + O(x^{-1/2}) \right).
\end{align*}
Therefore
$$
\frac{\ln N!}{N!}x^N = \left(1+O\left(\frac{1}{\ln x}\right)\right) [x \ln(x) e^x] \frac{e^{-\alpha^2/2}}{\sqrt{2\pi}} \frac{1}{\sqrt{x}}
$$
Now the sum of these $e^{-\alpha^2/2}{\sqrt{2\pi}} \frac{1}{\sqrt{x}}$ over $N \in [x-C \sqrt{x}, x+C \sqrt{x}]$, $(C$ fixed w.r.t. $x$), is a Riemann sum for the integral $\int_{-C}^C \frac{e^{-t^2/2}}{\sqrt{2\pi}}dt$. This integral can be arbitrarily close to 1 for $C$ large enough. 
It remains to be shown that for any $\epsilon >0$ (fixed w.r.t. $x$), 
$$
\sum_{n \notin [x-C \sqrt{x}, x+C \sqrt{x}]} \frac{\ln(N!)}{N!} x^N < \epsilon \, x \ (\ln x) \ e^x,
$$
for $C$ large enough. 
