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?

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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.

One may use, as $x \to 0$, the classic Taylor expansion:
$$\ln (1+x)=x-\frac{x^2}2+O(x^3)$$ giving, as $n \to \infty$,
$$n\ln \left(1-\frac1{\sqrt{n}}\right)=n \times\left(-\frac1{\sqrt{n}}-\frac1{2n}+O\left(\frac1{n^{3/2}}\right)\right)=-\sqrt{n}-\frac12+O\left(\frac1{\sqrt{n}}\right)$$ then, as $n \to \infty$,
$$\left(1-\frac {1}{\sqrt n}\right)^n=e^{-\sqrt{n}-1/2}\left(1+O\left(\frac1{\sqrt{n}}\right)\right) \to 0.$$