Let 
$$T(z)=\sum_{1}^\infty \frac{n^{n-1}x^n}{n!}.$$
This is known as the exponential generating function of rooted labeled trees, see for example https://math.berkeley.edu/~mhaiman/math172-spring10/trees.pdf. This function solves
$$T(x)=xe^{T(x)}.$$
Your sum is $y=T(1/e)$. So we have to solve the equation
$$y=e^{y-1}.$$
This has a root $y=1$ which is multiple (of multiplicity $2$).
This means that the graph of the LHS is tangent to the graph of the RHS
at the point $(1,1)$. Since the LHS is linear and RHS is convex,
our equation has unique solution, namely $y=1$, which proves your formula.