How can I get the result of this limits

I found a limits equation

$$\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^n=e^{-\lambda}$$

How can I get the result of $e^{-\lambda}$?

Normally, we can use

$$\lim_{x \to \infty}\left(1+\frac{n}{x}\right)^x=e^n$$

And how can I get $e^n$?

Answer

You may know that (sometimes this is used as definition of $e$)

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$
Taking $k$th powers, $k\in\Bbb N$, we obtain
$$e^k=\lim_{n\to\infty}\left(1+\frac1{n}\right)^{nk}=\lim_{n\to\infty}\left(1+\frac k{nk}\right)^{nk}.$$
The latter limit is the limit of a subsequence of $\lim_{n\to\infty}\left(1+\frac k{n}\right)^{n}$, hence this also converges to $e^k$, once we know it converges at all. In fact, the same method shows that more generally
$$\lim_{n\to\infty}\left(1+\frac {ak}n\right)^n =\left(\lim_{n\to\infty}\left(1+\frac {a}n\right)^n\right)^k$$
for $k\in\Bbb N$ and arbitrary $a$ (provided both limits exist).
As a consequence, $$\lim_{n\to\infty}\left(1+\frac {a}n\right)^n=e^a\qquad \text{for all }a\in\Bbb Q_{\ge0}.$$
Finally, using $(1-\frac1n)^n(1+\frac1n)^n=(1-\frac1{n^2})^n$, you can show that the same also hods for $a=-1$ and hence also for all $a\in\Bbb Q$.