Prove that $\phi(n_1 n_2 \cdots n_m)$ = $2^{m-1}\cdot \phi(n_1)\cdot\phi(n_2) \cdots \phi(n_m)$ where all of the $n$'s are distinct even perfect numbers.
I thought that because $\phi(2^k) = 2^{k-1}$ where $k$ is an positive integer, I should try to take out $2^m$ from all of the $n$'s, but I don't believe $2^m$ is a factor for all, and all of the $n$'s would have to be coprime after removing $2^m$. How would I go about getting the right side of the equation?
I think you were meant to assume that all the $n_i$ are even. There are no known, odd, perfect numbers.
Even perfect numbers are of the form $2^{p-1}(2^p-1)$ where $q=2^p-1$ is a Mersenne prime. See, e.g., https://proofwiki.org/wiki/Theorem_of_Even_Perfect_Numbers
Accordingly, write $n_i=2^{a_i}q_i$ where the $q_i$ are all distinct, prime, and odd. (we don't need the rest of the structure).
Then $$\varphi(\prod n_i)=2^{\sum a_i -1}\prod (q_i-1)$$
And $$\prod \varphi(n_i)=2^{\sum (a_i -1)}\prod q_i=2^{\sum a_i -m}\prod (q_i-1)$$
The desired equality follows by comparison.