Suppose that $X_i, i \in \mathbb{N}$ are integrable, independent random variables and $N:\Omega \to \mathbb{N}$ a random variable independent from all $X_i$. Does it hold that $\mathbb{E}[\prod_{i=1}^{N}X_i|N]=\prod_{i=1}^{N}\mathbb{E}[X_i|N]$?
I tried doing it by definition, but I get stuck at one point. After that, I tried to define $f(N, X_1, X_2, ...):=\prod_{i=1}^{N}X_i$, which is a measurable function, and then I tried to write $$\mathbb{E}[\prod_{i=1}^{N}X_i|N](\omega)=\mathbb{E}[f(N, X_1, X_2, ...)|N](\omega)=\mathbb{E}[f(N(\omega),X_1, X_2,...)]=\mathbb{E}[\prod_{i=1}^{N(\omega)}X_i]\overset{indep.}{=}\\=\prod_{i=1}^{N(\omega)}\mathbb{E}[X_i]\overset{indep.}{=}\prod_{i=1}^{N(\omega)}\mathbb{E}[X_i|N]$$ where the second equality is a naive generalization of $\mathbb{E}[f(X,Y)|\mathcal{F}](\omega)=\mathbb{E}[f(X(\omega),Y)]$ when $X$ is measurable w.r.t. $\mathcal{F}$ and $Y$ is independent from it, which is a well-known property of conditional expectations.
So, my questions are does the original statement hold and is this naive generalization correct?