How to prove Poisson Distribution is the approximation of Binomial Distribution?

Well, this is a basic fact of the exponential function $e^x$.

One definition of $e$ is the limit $\lim_{n\to\infty}(1+\frac1n)^n$. By a monotonicity argument one can prove $\lim_{x\to\infty}(1+\frac1x)^x=e$ where $x$ now ranges the real numbers.

Also note that $1-\frac1x=\frac{x-1}x=1/\frac x{x-1}=1/(1+\frac1y)=(1+\frac1y)^{-1}$ where $y=x-1$.
So, one has the following: $$\begin{aligned} \lim_{x\to\infty}(1-\frac1x )^x &= \lim_{y\to\infty}(1+\frac1y )^{-(y+1)} \\ &=\lim_{y\to\infty}(1+\frac1y)^{-y}\times\lim_{y\to\infty}(1+\frac1y)^{-1} \\ &=e^{-1}\times1=e^{-1}\,. \end{aligned} $$

From here, assuming $\lambda>0$, $$\begin{aligned} e^{-\lambda}=(e^{-1})^\lambda &= \lim_{x\to\infty}(1-\frac1x)^{\lambda x} &\to{\ z:=\lambda x} \\ &= \lim_{z\to\infty}(1-\frac\lambda z)^z\,. \end{aligned} $$

In consequence, we have this limit for every sequence $z_n\to\infty$ written in place of $z$ and limiting on the natural $n\to\infty$. In particular, this also holds for $z_n=n$.

Note that we had to take the turnaround for arbitrary real numbers instead of integers only because of the exponent $\lambda$.