integration - Expectation of $mathbb{E}(X^{k+1})$

I have difficulties with an old exam problem :

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. Show that

$$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\int_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$ Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

We have Fubini theorem, which we can apply to a $\mathbb{B}(\mathbb{R})\otimes\mathcal{F}$-measurable function because the Lebesgue measure is $\sigma$-finite and $\mathbf{P}$ is also $\sigma$-finite because it is a probability. I think we can write $\mathbf{P}(X\geq t)$ as $\int_{\{ X(w)\geq t\}} d\mathbf{P}(\omega)$ but I don't know how to proceed next. Especially I don't see how to introduce the $\int_{\Omega}$.

Edit

From the comments, there must be an error in the description of the exam problem. It should have been the following :

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. Show that

$$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\mathbf{1}_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$ Where $\mathbf{1}_{\{X(\omega)\geq t\}}$ is the characteristic function of $\{ X(\omega)\geq t\}$

Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

Answer

By Fubini's theorem, we have

$$\begin{align*} \int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt&= \int_0^\infty t^k \mathbb E\left[1_{\{\omega:X(\omega)\geqslant t\}} \right]\mathsf dt\\ &=\int_0^\infty t^k\int_{\Omega} 1_{\{\omega:X(\omega)\geqslant t\}}\mathsf d\mathbb P\;\mathsf dt\\ &=\int_{\Omega}\int_0^{X(\omega)}t^k \mathsf dt\; \mathsf d\mathbb P\\ &=\int_{\Omega} \frac1{k+1}X^{k+1}(\omega)\mathsf d\mathbb P(\omega)\\ &= \frac1{k+1}\mathbb E[X^{k+1}]. \end{align*}$$

Hence

$$\mathbb E[X^{k+1}] = (k+1)\int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt.$$

The crucial part here is that

$$1_{\{\omega : X(\omega) \geqslant t\}}(\omega) = 1_{\{t: t\leqslant X(\omega)\}}(t).$$