Euler's identity is usually stated as

e^{iπ} = −1

Rewriting slightly, we have e^{iπ} + 1 = 0. This succinct result relating the fundamental concepts of e, i, π, 1, and 0, is a special case of the trigonometric identity e^{iθ} = cos θ + i sin θ, where θ = π.

To establish the identity for all values of θ, we use the Maclaurin power series:

e^{z} = 1 + z/1! + z^{2}/2! + z^{3}/3! + z^{4}/4! + ...

It follows that

e^{iθ} = 1 + iθ/1! − θ^{2}/2! − iθ^{3}/3! + θ^{4}/4! + iθ^{5}/5! − θ^{6}/6! − iθ^{7}/7! + ...

e^{iθ} = 1 − θ^{2}/2! + θ^{4}/4! − θ^{6}/6! + ... + i(θ/1! − θ^{3}/3! + θ^{5}/5! − θ^{7}/7! + ...)

e^{iθ} = cos θ + i sin θ

Note: One may also establish the result using differential equations:

Let f(θ) = e^{iθ} and g(θ) = cos θ + i sin θ

Observe that f'(θ) = i f(θ) and that g'(θ) = −sin θ + i cos θ = i^{2} sin θ + i cos θ = i (cos θ + i sin θ) = i g(θ)

Finally, f(0) = 1 = g(0).