Problem calculating mean of probability distribution

I have a discrete probability distribution where $$P(X=x)=(Factorial[n]/(n^x))*StirlingS2[x-1,n-1]$$. So, when trying to calculate its mean with the following property: $$\mu = \sum_{i} x_i \cdot P(x_i)$$, I insert this expression in Mathematica:

Sum[(Factorial[n]/(n^x))*StirlingS2[x - 1, n - 1]*x, {x, 0, Infinity}]

But the program doesn't output the correct answer, which in this case would be n*HarmonicNumber[n], instead it provides: $$\sum _{x=0}^{\infty } x n! n^{-x} \mathcal{S}_{x-1}^{(n-1)}$$ as output.

I have tried to encapsulate the operation inside FunctionExpand and FullSimplify but I can´t make it work.

$$\mathcal{S}_{x-1}^{(n-1)}=\frac{\sum _{j=0}^{n-1} (-1)^{j+n-1} \binom{n-1}{j} j^{x-1}}{(n-1)!}$$
FullSimplify[Sum[Sum[(Factorial[n]/(n^x))*((1/(n - 1)!) *(-1)^(j + n - 1)  Binomial[n - 1, j]  j^(x - 1))*x // FullSimplify, {x, 0, Infinity}] // FunctionExpand, {j, 0, n - 1}], Assumptions -> {n \[Element] PositiveIntegers}]

• Thanks for the solution, is there any way to natively use this definition in the StirlingS2[] function? Commented Apr 14 at 10:48