Moments and non-negative random variables? I want to prove that for non-negative random variables with distribution F:
$$E(X^{n}) = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx$$
Is the following proof correct?
$$R.H.S = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx = \int_0^\infty n x^{n-1} (1-F(x)) dx$$
using integration by parts:
$$R.H.S = [x^{n}(1-F(x))]_0^\infty + \int_0^\infty x^{n} f(x) dx = 0 + \int_0^\infty x^{n} f(x) dx = E(X^{n})$$
If not correct, then how to prove it?

Here's another way.  (As the others point out, the statement is true if $E[X^n]$ actually exists.)
Let $Y = X^n$.  $Y$ is non-negative if $X$ is.
We know 
$$E[Y] = \int_0^{\infty} P(Y \geq t) dt,$$
so
$$E[X^n] = \int_0^{\infty} P(X^n \geq t) dt.$$
Then, perform the change of variables $t = x^n$.  This immediately yields 
$$E[X^n] = \int_0^{\infty} n x^{n-1} P(X^n \geq x^n) dx = \int_0^{\infty} n x^{n-1} P(X \geq x) dx.$$
