Complex number \$e^{jk\pi} = (-1)^k\$
There is a very famous identity, know as Euler's identity, which says:

$$e^{i\pi}+1=0 \rightarrow e^{i\pi}=-1$$

It's an identity which should be known by all engineers and mathematicians, and you can simply substitute it in when needed.

As to your question

$$a^{bc} = (a^b)^c$$

so

$$e^{ik\pi}=(e^{i\pi})^k=(-1)^k$$



In case you are wondering, and as described in full on the Wiki article in the above link, this comes from the Euler's formula:

$$e^{ix} = \cos x + i\sin x$$

What happens when you set \$x = \pi\$?

Well, you get:

$$e^{i\pi} = \cos (\pi) + i\sin (\pi) = -1 + 0i = -1$$