This looks basically correct to me. The part you're worried about can be made a little more precise if you like, as follows:

The statement you're actually trying to prove is

∀n∈ℕ∀p∈{p∈ℕ:0≤p≤n}: nCp∈ℕ.

You prove this by induction on n, so the proposition in question is

P(n) = "∀p∈{p∈ℕ:0≤p≤n}: nCp∈ℕ".

Your base case is fine. At the induction hypothesis, you simply assume P(n). This isn't strong induction; you only need to assume a single case.

Then at the induction step, you use P(n) twice: once to show that nC(p-1) is natural, and once to show that nCp is.

Incidentally, there is one more nitpick you have to worry about: when p=n+1, you can't use the induction hypothesis because nC(n+1) doesn't even make sense. Instead, you have to handle this case separately by showing that (n+1)C(n+1) is natural.