the question is to prove $\binom{n}{1}$ is always a natural number by induction. I'm able to understand $\binom{1}{1}$. Then I though I'm supposed to assume $\binom{n}{p}$ is a natural number and prove for $\binom{n+1}{p+1}$.
My first doubt is why prove for $\binom{n+1}{p}$ instead of $\binom{n+1}{p+1}$
Secondly even for $\binom{n+1}{p}$ I'm not able to understand the proof. We have only assumed that $\binom{n}{p}$ is a natural number. By what means do we know that $\binom{n}{p-1}$ is a natural number? If I know how $\binom{n}{p-1}$ is a natural number I can simply use closure due to addition to say $\binom{n+1}{p}$ is a natural number if I'm correct.
For fixed $n$, let $A(n)$ be the assertion $$ \binom{n}{p}\text{ is a natural number for all } p \text{ such that }1\leq p\leq n $$ In order to prove that $A(n)$ is true for all $n$, you proceed by induction on the variable $n$.
Base case: When $n=1$ all you have to verify is that $\binom{1}{1}$ is a natural number. Since $\binom{1}{1}=1$, this is true.
Inductive step: We suppose that $A(n)$ is true for some given $n$ and show that $A(n+1)$ must follow.
That is, we must show that $\binom{n+1}{p}$ is a natural number for all $p$ such that $1\leq p\leq n+1$.
We will consider three (exhaustive) cases. Note that we will not make use of the induction hypothesis for the first two.
All in all, we have proved that $\binom{n+1}{p}$ is a natural number for all $p$ such that $1\leq p\leq n+1$ if $\binom{n}{p}$ is a natural number for all $p$ such that $1\leq p\leq n$. Equivalently, we showed that $A(n)$ implies $A(n+1)$. This concludes the inductive step and the proof by induction on $n$.