I had a statement in a lecture which says that if $X$ is $\mathcal G$-measurable and $X$ and $XY$ are integrable random variables then this equality holds: $\mathbb E(XY|\mathcal G) = X\mathbb E(Y|\mathcal G)$
Can we also say form the above statement that these equalities hold:
$\mathbb E(X^n|X) = X^n$ (here we use the theorem $n-1$ times)
$\mathbb E(X^nY|X) = X^n \mathbb E(Y|X)$ (the same as above)?