This is problem 3(b) from Spivak's Calculus 4th ed., Chapter 2:
Prove that $\binom{n}{k}$ is always a natural number if $0\leq k\leq n$
This is my proof:
If $n=0, \binom{0}{0}= \frac{0!}{0!}=1$
$\therefore \binom{0}{0} \in \Bbb N$, so the statement holds true for $n=0$.
Assuming that the statement holds true for $n=m$ such that $$\binom{m}{0},\binom{m}{1},\binom{m}{2},...,\binom{m}{m-1},\binom{m}{m} \in \Bbb N$$
Then for $n=m+1$, it follows that $\binom{m+1}{k}\in\Bbb N$ since $\binom{m+1}{k}=\binom{m}{k-1}+\binom{m}{k}$ and $\binom{m}{k-1}$ or $\binom{m}{k}\in \Bbb N$ by assumption.
$\therefore$ It has been proven that if the statement holds true for $n=m$ then it must hold true for $n=m+1 \Rightarrow$ By the principle of mathematical induction, the statement must hold true for all $0\leq k\leq n$ and $k,n\in \Bbb Z^{\geq}$. I'm posting this proof as it is different to the one Spivak provides and I'm not sure if it makes sense. Is it correct?
In your proof the case of $ k =m +1$ has to be addressed separately.
Otherwise it is a good proof.