Replacing $n!$ with Stirling's approximation in $e^x = \sum_n \frac{x^n}{n!}$

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I was wondering if there is a closed-form expression for

$$\sum_{n=0}^{\infty} \frac{x^n}{e^{-n}n^n},$$

although I expect there is none because Mathematica cannot compute it. However, from Stirling's approximation

$$n! \approx e^{-n} n^n$$

I would expect this sum to be $\approx e^x$ and, in particular, convergent everywhere.