Spivak's calculus, chapter 2 question 3 (b) 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.

 A: 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.


*

*If $p=1$, then $\binom{n+1}{p}=\binom{n+1}{1}=n+1$ is a natural number.

*If $p=n+1$, then we have $\binom{n+1}{p}=\binom{n+1}{n+1}=1$ which is
a natural number.

*If $2\leq p\leq n$, then $p-1$ and $p$ are between $1$ and $n$
inclusively. Making use of the known identity $$
   \binom{n+1}{p}=\binom{n}{p-1}+\binom{n}{p}\tag{$\star$} $$ we can now
use the assumed truth of $A(n)$. $A(n)$ says, in particular, that
both terms in the right-hand side of $(\star)$ are natural numbers
and since the sum of two natural numbers is a natural number, we see
that $\binom{n+1}{p}$ is also a natural number.


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 $n$.
