To prove $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is an integer, use Mathematical Induction on $k$
Base Step: $\binom{n}{0}=\binom{n}{n}=1$
Inductive Step: Assume that $\binom{n}{k}$, $k=1,2,...,n-1$ are all integers. We need to show that $\binom{n+1}{k}$, $k=1,2,...,n$ are all integer
Given $k=1,2,...,n$
$\binom{n+1}{k}=\frac{(n+1)!}{k!(n-k+1)!}$
$=\frac{n!}{(k-1)!(n-k)!}\frac{n+1}{(n-k+1)k}$
$=\frac{n!}{(k-1)!(n-k)!}(\frac{1}{(n-k+1)}+\frac{1}{k})$
$=\frac{n!}{(k-1)!(n-k+1)!}+\frac{n!}{k!(n-k)!}$
$=\binom{n}{k-1}+\binom{n}{k}$
Is this a complete and correct proof ? If not, what is wrong with it ?