In my book has the follow :
For all $n \in\mathbb{N^*}$ and $ i \in\mathbb{N}$ with $0 \le i \le n$. We have this:
$$\binom{n}{i} + \binom{n}{i+1} = \binom{n+1}{i+1}$$
For $i = n$, the book said is easy to proof, after I have asked in here, I understand because for to 1 this proof is simple, how like in the comments is because of the 0 $\le$ i $\le$ n.
Then, when the $i = n$:
\begin{align} \binom{n}{n} + \binom{n}{n+1} &= \frac{n!}{n!(n-n)!} + \frac{n!}{n+1!(n-(n+1))!}\\ = 1\end{align}
because i > n , is defined $\binom{n}{i} = 0.$
First of all notice that $$\binom{n}{i} + \binom{n}{i+1} = $$
by definition $$ \frac {n!}{i!(n-i)!}+ \frac {n!}{(i+1)!(n-i-1)!} =$$
Common denominator $$ \frac {n!(i+1+n-i)}{(i+1)!(n-i)!}=$$
Simplify $$\frac {(n+1)!}{(i+1)!(n-i)!}=$$
by definition $$ \binom{n+1}{i+1}.$$