How can i prove this? (I don't know anything about the stirling numbers, because i'm currently seeing this kind of moments) $$\mu_r'(X)=\sum_{k=0}^r S_r^{(k)} \mu_{[k]}(X)$$ where the stirling numbers of second class are denoted by $S_x^{(s)}$
Thank you in advance.
The Stirling numbers satisfy the polynomial identity $$x^r = \sum_{k=0}^r S_r^{(k)} (x)_k$$ where $(x)_0 = 1$ and for $k \ge 1$, $(x)_k = x(x-1)(x-2)\cdots(x-k+1)$.
In fact, you could take this as one definition of the Stirling numbers: since the polynomials $\{(x)_0, (x)_1, \dots, (x)_r\}$ include a polynomial of each degree $0$ through $r$, we can write any polynomial of degree at most $r$ as a linear combination of these polynomials. In particular, we can do this for $x^r$. The Stirling numbers are the coefficients we get if we do this.
Anyway, from the polynomial identity above, you should be able to obtain the moment identity by replacing $x$ with your random variable, then taking the expectation of both sides.