let $\mu$ = $E[V]$ where V is a typical family resulting from one individual, then prove that $E[X_n] = \mu^2$

Question

I used the fact that $G_{v}^{'}(1) = \mu$ , $G_v(1) = 1$ and $G_n(z) = G_{n-1}(G_v(z))$

I then got the following: $E[x_n] = G_{v}^{'}(1) = \frac{d}{dv} G_{n-1}(G_v(1))$

which i simplified to: $G_{n-1}(G_v(1))G_{v}^{'}(1)$

which $= G_{n-1}(1)\mu.$ But I am not sure what to do next.