We observe from Pascal's triangle that
$$\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$$
We prove this statement correct by expanding the right side:
$$\frac{n!}{(k-1)!(n-k+1)!}+\frac{n!}{(k)!(n-k)!}\rightarrow\frac{(n+1)!}{k!(n-k+1)!} $$
Next we assume that it is also true for some number $n+1$:
$$\binom{n+1}{k-1}+\binom{n+1}{k}$$
This is equivalent to:
$$\binom{n+1}{k-1}+\binom{n}{k-1}+\binom{n}{k}$$
Expanding we get:
$$\binom{n+1}{k-1}+\binom{n}{k-1}+\binom{n}{k}\rightarrow\binom{n+2}{k}$$
Which should make the proof complete since we can assert that because of this condition: $0\le k\le n$, and by proving the equality below, all other numbers, being recursively defined, are also natural numbers.
$$\binom{1}{0}=\binom{1}{1}=1$$