Proof $\lim\limits_{n \rightarrow \infty}n(a^{\frac{1}{n}}-1)=\log a$ I want to show that for all $a \in \mathbb{R }$
$$\lim_{n \rightarrow \infty}n(a^{\frac{1}{n}}-1)=\log a$$
So far I've got $\lim\limits_{n \rightarrow \infty}ne^{(\frac{1}{n}\log a)}-n$, but when i go on to rearrange this, i come  after a few steps back to the beginning...
We have no L'Hôpital and no differential and integral calculus so far.

Another way is to notice that this is a definition of derivative (if you are allowed to use it):
$$
\lim_{v \to 0}\frac{a^v-1}{v} = a^v \log a\Bigg|_{v=0}=\log a
$$
