An exponential limit

Question

According to Wolfram Alpha, for $a > 0$ $$\lim \limits_{n \to \infty} \left( 1 - e^{na} \right)^{1/n} = e^a.$$ Is this easy to prove?

EDIT: Actually, in the case I need, $a < 0$. If I understood correctly this is covered by the answers below.

Answer 1

Some algebra:

$$(1-e^{na})^{1/n}=e^a(-1)^{1/n}(1-e^{-na})^{1/n}$$

As $n\to\infty$, note that

$$(-1)^{1/n}=e^{\frac1n(2k+1)\pi i}\to\begin{cases}+1&;n\equiv0\pmod2\\-1&;n\equiv1\pmod2\end{cases}$$

The second limit:

$$\lim_{n\to\infty}(1-e^{-na})^{1/n}=(1-0)^0=1$$

Thus,

$$\lim_{n\to\infty}(1-e^{na})^{1/n}=e^a\times1\times\pm1=\pm e^a$$

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If, however, you stick to one branch and don't jump branches as $n\to\infty$, then

$$\lim_{n\to\infty}(-1)^{1/n}=1^0=1$$

And thus, we get the different result of

$$\lim_{n\to\infty}(1-e^{na})^{1/n}=e^a$$

Answer 2

If we remain in the reals, for even $n$ the terms cease to exist after a short while, while for odd $n$,

$$\lim_{n\to\infty}(1-e^{na})^{1/n}=\lim_{n\to\infty}(e^{-na}-1)^{1/n}e^a=-e^a.$$

(But WA doesn't stay in the reals.)

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Update:

For $a<0$, the expression tends to $1^0$.