Calculate $\mathbb{E}(W_t^k)$ for a Brownian motion $(W_t)_{t \geq0}$ using Itô's Lemma

Show by using Ito's Lemma, for $k \geq 2$ the following result hold.

$$E[W(t)^k] = \frac{1}{2} k(k-1)\int_0^t E[W(s)^{k-2}]ds$$

where $W(t) = N(0,t)$ is standard Brownian motion.

I think $E[W(t)^k]$ is an expectation over space,

$$E[W(t)^k] = \int_x x^k N(0,t) dx $$

How could I derive to the R.H.S. which contains a time integral? I am lacking a hint on this.

Let $f(x) := x^k$, then by Itô's formula

$$W_t^k = \int_0^t k \cdot W_s^{k-1} \, dW_s + \frac{1}{2} k \cdot (k-1) \cdot \int_0^t W_s^{k-2} \, ds$$

Since $(t,\omega) \mapsto \left(\int_0^t k \cdot W_s^{k-1} \, dW_s \right)(\omega)$ is a martingale, we have $$\mathbb{E}(W_t^k) = 0+ \frac{1}{2} k \cdot (k-1) \cdot \mathbb{E} \left( \int_0^t W_s^{k-2} \, ds \right) = \frac{1}{2} k \cdot (k-1) \cdot \int_0^t \mathbb{E}(W_s^{k-2}) \, ds$$