A case of double induction? The question being: Given $n,m\in\mathbb{Z}$ and for $n>0,m\ge-1$ prove that $1+nm=(1+m)^n$. I have never laid eyes upon a problem of inducting across two variables, but I assume that is the case here? If so, then how does one proceed in a manner that completely covers all cases? It is easy enough to fix one variable and induct on the other, but how does one go about doing both at the same time? I am not familiar with any canonical forms to this type of question. Regarding the sufficient effort clause of the homeworks tag; attempts have been made that involve fixing one, proving for all cases of the other, and then attempting to "iterate" forward, but that has proven rather haphazard. Am I over complicating the question entirely?

If you set the proof up as an inductive proof on one variable, and then, during the inductive step for that variable, conduct an inductive proof on the second variable, I think you'll find that everything proceeds exactly the way you would expect it.

I.e., your base case might be $n=1$, then your outermost inductive step would be to assume the statement is true for $n=k$ and attempt to prove it for $n=k+1$. Then, you would attempt to prove the statement with $n$ replaced by $k+1$ by induction on $m$. Does that make sense?

However, I should point out that the statement you're trying to prove appears to be false.