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Prove that $e^{i\pi z}=(-1)^z$ for all $z\in\mathbb{C}$

For a long time I believed that following reasoning is correct $e^{i\pi z}=(e^{i\pi})^{z}=(-1)^z$. Recently I discovered that $ e^{ab}=(e^a)^b$ is not true in general case. Could you provide me any hints?

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    What's your definition of $(-1)^z$? – 2017-02-08
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    what is your reasoning behind $e^{ab} = (e^a)^b$ being not generally true - is that definitely right? – 2017-02-08
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    You know that for $z \in \mathbb{C}$, $\log z$ is well-defined only up to $2ik \pi, k \in \mathbb{Z}$. Then $e^{\frac{\log z}{2}} = \ ?$. It is the same here, except that when $1/2$ is replaced by a complex number $b$, then $b 2i k \pi$ and $e^{b\log z}$ can take infinitely many values – 2017-02-08
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    For general $a \in \Bbb C$, we define $a^z := \exp (z \log a)$ (for a choice of branch cut of $\log$). Now, the possible values of $\log (-1)$ are the solutions to $\exp w = -1$, namely, $(2 \pi k + 1) i$, $k \in \Bbb Z$. – 2017-02-08
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    Thank you for your help. Here's reasoning that satisfies me. Because of definition $ a^b=\exp(b \log(a))$ for all $a,b\in \mathbb{C}$, I have $(-1)^z=\exp(z\log(-1))$. Using Euler's formula I get $\log(-1)=\pi i$. Summarising $(-1)^z=\exp(z \pi i)$ @Cato Note that $e^{2\pi i \sqrt{2}}\neq (e^{2\pi i})^\sqrt{2}$ – 2017-02-08

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