Apparent Contradiction by WolframAlpha in Calcuation of $\sum_{k=1}^\infty k^{k-1} e^{-k} / k!$

The original sum is exactly 1. The numeric approximation of the last sum in your question is actually $$1.0017976085327493751120130443778945039608311925192\ldots,$$ which is greater than $1$, so that particular input was not evaluated properly.

I leave it as an exercise to show that $$\sum_{k=1}^\infty \frac{k^{k-1} e^{-k}}{k!} = 1.$$

Incidentally, the sum arising from the coarser form of Stirling's approximation is $\zeta(3/2)/\sqrt{2\pi} \approx 1.04219$.

Lambert's W function has series $$W(x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}\,k^{k-1}}{k!}x^k \tag 1$$

which has radius of convergence $1/e$. The series in $(1)$ can be obtained using Lagrange's Inversion Theorem.

Since we have $-W(-1/e)=1$, then from $(1)$ we see that

$$\sum_{k=1}^\infty \frac{(-1)^{k-1}\,k^{k-1}}{k!}\left(-\frac1e\right)^k=1$$

from which we recover the identity

$$\sum_{k=1}^\infty \frac{k^{k-1}e^{-k}}{k!}=1$$