Nice equality involving summation, factorials, and $e$ I came across this nice equality that I don't know how to solve:
$$\sum_{j=1}^{\infty} \frac{e^{-j}j^{j-1}}{j!}=1.$$
I'm guessing the first step is to write out the Taylor series, which gives a double sum of:
$$\sum_{j=1}^{\infty} \frac{\left(\sum_{n=1}^{\infty}\frac{(-j)^n}{n!}\right)j^{j-1}}{j!}$$
and maybe this can be seen to be the product of two other functions in Taylor form, which I can't see. Also, I know this can be solved by elementary methods, so a "simple" solution would be nice. Help is appreciated!

Hint. For $|x|< 1/e$, the power series 
$$T(x)=\sum_{j=1}^{\infty}\frac{j^{j-1}}{j!}\,x^j,$$
which is the exponential generating function for rooted labeled trees,
satisfies the identity
$$T(x)=xe^{T(x)}.$$
Now you need to show that $T(1/e)=1$.
