Show that ${{n}\choose{x}}p_n^x(1-p_n)^{n-x}=e^{-\lambda}\frac{\lambda^x}{x!}$ 
Question: Suppose $x $ is a non negative integer. Define ${{m}\choose {x}}=0$ if $x>m $. Let $\{p_n\}$ be sequence satisfying  $0 <p_n <1$ and $\lim\limits_{n\to\infty} np_n=\lambda$. Show that $${{n}\choose {x}}p _n^x (1-p_n)^{n-x}=e^{-\lambda}\frac{\lambda^x}{x!}$$

Is this equivalent to the proof of Poisson Distribution formula ? I am asking this because in the statent ment of Poisson Distribution formula $np$ is constant but here when $n\to\infty $ $np\to $some constant$=\lambda $. Also in the Poisson Distribution Formula $\lim\limits_{n\to\infty} {{n}\choose{x}}p_n^x(1-p_n)^{n-x}=e^{-\lambda}\frac{\lambda^x}{x!}$ but we have to prove for any $n $ there is no limits. So is the proof for the problem and the proof of the Poisson distribution formula same ?
Note: There is no limit in the formula in the problem. We have to prove $${{n}\choose {x}}p _n^x (1-p_n)^{n-x}=e^{-\lambda}\frac{\lambda^x}{x!}$$ not $$\lim\limits_{n\to\infty} {{n}\choose{x}}p_n^x(1-p_n)^{n-x}=e^{-\lambda}\frac{\lambda^x}{x!}$$
 A: The limit you have written is the formal statement of the Poisson limit theorem.
The version you saw before has a slightly less general assumption (it forces $np_n = \lambda$ for all $n$, rather than $np_n \to \lambda$). The proofs will be very similar, but you'll probably have to do something extra for the more general claim.
In both statements, there is a limit as $n \to \infty$; I am not sure what you mean by "we have to prove for any $n$ there is no limits."
A: $\boxed{\text{Hint}}$
$$\binom{n}{x}p_n^x(1-p_n)^{n-x}=\frac{n!}{x!(n-x)!}\frac{(np_n)^x}{n^x}\frac{(1-np_n/n)^n}{(1-p_n)^x}$$
Since $\lim_{n\rightarrow\infty}np_n=\lambda$, we get $$\boxed{\lim_{n\rightarrow\infty}(1-np_n/n)^n=\lim_{n\rightarrow\infty}(1-\lambda/n)^n=e^{-\lambda}}$$
$$\boxed{\lim_{n\rightarrow\infty}(np_n)^x=\lambda^x}$$ and try to show that $$\boxed{\lim_{n\rightarrow\infty}\frac{n!}{(n-x)!n^x}=1\\\lim_{n\rightarrow\infty}(1-p_n)^x=1}$$
A: For fixed $x$,$$\frac{\binom{n}{x}}{n^x/x!}=\prod_{i=0}^{x-1}(1-i/n)=\exp\sum_{i=0}^{x-1}\underbrace{\ln(1-i/n)}_{\sim-i/n}\approx\exp\frac{-x(x-1)}{2n}\stackrel{n\to\infty}{\to}1.$$As $n\to\infty$, $1-p_n\to1$ so$$\binom{n}{x}p_n^x(1-p_n)^{n-x}\sim\frac{\left(\frac{np_n}{1-p_n}\right)^x(1-p_n)^n}{x!}\sim\frac{\lambda^x e^{-np_n}}{x!}\stackrel{n\to\infty}{\to}\frac{\lambda^x e^{-\lambda}}{x!}.$$
