How come e^πi=-1?

Question

I enjoy math as a student but I still can't come up with an answer as to why this is true.

Answer 1

One answer is that e^θi = cosθ + i sinθ . When θ=π, cosθ=-1 and sinθ=0

Do the infinite series expansion e^x = 1 + x^1/1! + x^2/2! + ..., when x=θi . The odd-order terms all have i in them, even-order terms all lack i. Separate the terms into with- and without-i, and you'll see the sin and cos expansions.