Let $G$ be any group with $|G| =N=123456789$. Let $g\in G$, and $M=2^{2^{2524820221019}}$. Then $g$ satisfies $g^M = e$ if and only if $g = e$. 
Let $G$ be any group with $|G| = N = 123456789$. Let $g$ be any element of $G$, and $M = 2^{2^{2524820221019}}\ \ \ \ \ $. Then $g$ satisfies $g^M = e$ if and only if $g = e$.

I am trying to determine if this is true. I think information like $M$ is even and $N$ is odd is useful. I think it is false since $N$ cannot divide $M$ but other than that I don't really understand what is going on.

Since M only has factors of 2, the only factor of M that is odd is 1. Therefore $g^M$ = e implies $g^1$ = e implies g = e.
