Suppose $M_n \rightarrow M$ in $L^1$ norm, then $\mathbb{E}[\mathbb{E}[M_X(t)^{M_n} | M_n]] \rightarrow \mathbb{E}[\mathbb{E}[M_X(t)^{M} | M]]$?

Question

Suppose we have two r.v. $M, X$ and some sequence of r.v. $(M_n)_n$ s.t. $M_n \rightarrow M$ in $L^1$ norm, i.e. $\mathbb{E}[|M_n - M|] \rightarrow 0$. I would now like to show that we also have: $$ \mathbb{E}[\mathbb{E}[M_X(t)^{M_n} | M_n]] \rightarrow \mathbb{E}[\mathbb{E}[M_X(t)^{M} | M]], $$ here $M_X(t)$ denotes the moment generating function of $X$ at $t$. I would guess that we can replace $M_X(t)$ by any value $x$ so we would like to show that for arbitrary $a$ we have: $$ \mathbb{E}[\mathbb{E}[a^{M_n} | M_n]] \rightarrow \mathbb{E}[\mathbb{E}[a^{M} | M]], $$ and we need to find a good bound on $$ \mathbb{E}[\mathbb{E}[a^{M_n} | M_n] - \mathbb{E}[a^{M} | M]] $$ which does not seem too hard, but I don't really know how to get started.