For binomial n,p choosing k:

$$\mathbb{P}(X=k)={n \choose k}p^k(1-p)^{n-k}\ \ \ \ k=0,1,...,n$$

Taking %%n\to\infty%% and %%p=\frac{\lambda}{n}%%:

$$\lim_{n\to\infty}{n \choose k}p^k(1-p)^{n-k}=\lim_{n\to\infty}\frac{n!}{k!(n-k)!}\Big(\frac{\lambda}{n}\Big)^k\Big(1-\frac{\lambda}{n}\Big)^{n-k}$$

Moving constants to front and splitting third term:

$$=\Big(\frac{\lambda^k}{k!}\Big) \lim_{n\to\infty} \frac{n!}{(n-k)!}\Big(\frac{1}{n^k}\Big) \Big(1-\frac{\lambda}{n}\Big)^n \Big(1-\frac{\lambda}{n}\Big)^{-k}$$

Considering first two terms

$$\lim_{n\to\infty} \frac{n!}{(n-k)!}\Big(\frac{1}{n^k}\Big) =$$ $$\lim_{n\to\infty} \frac{n(n-1)(n-2)...(n-k)(n-k-1)...(1)}{(n-k)(n-k-1)...(1)}\Big(\frac{1}{n^k}\Big)$$

Bottom cancels with top to leave k terms of the top. Equivalent to each term of the top divided by n, which all approach 1 as n tends to infinity.

$$=\lim_{n\to\infty} \frac{n(n-1)(n-2)...(n-k+1)}{n^k}=1$$

For the second part, we note that it shares the structure of the definition of %%e%%.

$$\lim_{n\to\infty} \Big(1-\frac{\lambda}{n}\Big)^n$$

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

Taking %%x=-\frac{n}{\lambda}%%

$$\lim_{n\to\infty} \Big(1-\frac{\lambda}{n}\Big)^n = \lim_{n\to\infty} \Big(1+\frac{1}{x}\Big)^{-\lambda x} = e^{-\lambda}$$

Finally, the third part goes to 1 as n goes to infinity.

$$\lim_{n\to\infty} \Big(1-\frac{\lambda}{n}\Big)^{-k}=1$$

Giving us, for %%p=\frac{\lambda}{n}%%

$$\lim_{n\to\infty}{n \choose k}p^k(1-p)^{n-k}=\frac{\lambda^ke^{-\lambda}}{k!}$$