probability - 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?

Answer

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

$$E[X^n] = \int_0^{\infty} n x^{n-1} P(X^n \geq x^n) dx = \int_0^{\infty} n x^{n-1} P(X \geq x) dx.$$