$B$ is a brownian motion.
Let $n$ be an integer larger than 1, by applying Ito's formula to $f(t, x) = x^{2n}$ we find that : $$ \forall t \geq 0, B_{t}^{2 n}=2 n \int_{0}^{t} B_{s}^{2 n-1} d B_{s}+n(2 n-1) \int_{0}^{t} B_{s}^{2 n-2} d s $$
since stochastic integrals have mean $0$ then by taking the expectation of both sides and using fubini's theorem we get :
\begin{align} \mathbb{E} [B_{1}^{2n}] &= n(2n-1) \mathbb{E} [\int_{0}^{1} B_{s}^{2 n-2} d s ] \\ &=n (2n-1) \int_{\Omega}\int_{0}^{1} B_{s}^{2 n-2} d s d \mathbb{P} = n(2n-1)\int_{0}^{1}\mathbb{E}[B_s^{2n-2}] ds \end{align}
so somehow, we have to prove that :$n\int_{0}^{1}\mathbb{E}[B_s^{2n-2}] ds =\mathbb{E}\left(B_{1}^{2 n-2}\right) $ using Ito's calculus
I tried using the expected value formula as the integral of the survival function but it led me to nowhere
how to continue ?