We know $e^{2\pi i} = 1$, and that $(x^m)^n = x^{mn}$. This way, we can rewrite $e^{n}$ as some version of $(e^{2\pi i})^{\frac{n}{2\pi i}}$ for most n (right?).
But if this is true, then why isn't $e^3 = 1$, for example, if we can rewrite it as $(e^{2\pi i})^{\frac{3}{2\pi i}} = (1)^{\frac{3}{2\pi i}} = 1$ ? What am I missing here?
I just came upon this issue by accident while doing a problem, and I'm not sure how to best resolve it.
Good question! The answer is that although the rule ${(x^b)^c} = x^{bc}$ holds when $b$ and $c$ are integers, it does not hold in general when they are not integers.
Consider the following simpler example. As you know, $(-1)^2 = 1$. Raising both sides to the power $\frac12$, we get $${((-1)^2)}^{1/2} = 1^{1/2},$$ which is still correct, but we cannot then apply the ${(x^b)^c} = x^{bc}$ rule to the left side to obtain $$(-1)^{2\cdot(1/2)} = (-1)^1 = -1 = 1.$$