I want to prove that
$$e^{jn\pi} = (-1)^n e^{-jn\pi} \quad (1)$$
My attempt
$$ e^{jn\pi} = (-1)^n e^{-jn\pi} \iff ln(e^{jn\pi}) = ln((-1)^n) + ln(e^{-jn\pi} ) \iff jn\pi = jn\pi - jn\pi$$ $$\iff n = 0 $$ $$\text{But this should be true } \forall n, \text{therefore something went wrong here}$$
I am very confident that there is a more beautiful proof based on Euler's formula but I can't find it. What I've also tried is:
$$e^{jn\pi} = cos(n\pi) + j sin(n\pi) = (-1)^n + j sin(n\pi) = ? $$
The statement is false. Notice that
$$(e^{j \pi})^n = (-1)^n$$
Then your statement is
$$(-1)^n = (-1)^n (-1)^{-n}$$
However, the right-hand-side simplifies to $(-1)^0 = 1$, so the statement can only be true for $n$ even at best. For instance, letting $n=3$ produces the statement $-1 = 1$, though at least $n=4$ is true.