Elementary question about the limit $\big( 1 - \frac 1 {\sqrt n}\big )^n$, $n\to\infty$. When calculating the limit $L=\big( 1 - \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:
$$
L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}
$$
As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.
It's like we're distributing the parenthesis somehow:
$$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$
The question is: Why can we do this? Which property are we using?

You are looking for the following statement.
Let $(a_n)$ be a converging sequence of non-negative real numbers with $\lvert \lim_{n → ∞} a_n \rvert < 1$. Furthermore, let $(e_n)$ be an unbounded, increasing sequence. Then $({a_n}^{e_n})$ converges to zero.
Proof idea. Let $a = \lim_{n → ∞} a_n$. Let $K > 1$ be a real number. Then $({a_n}^K)$ converges to $a^K$ by limit theorems. Since $e_n > K$ for large enough $n$, almost all members of $({a_n}^{e_n})$ are smaller than $a^K$, so both limes inferior and limes superior of the sequence $({a_n}^{e_n})$ lie within the interval $[0..a^K]$.
Because $K$ was arbitrary and $\lvert a \rvert < 1$, the sequence must converge to zero.
