show $\sum_{k=0}^n {k \choose i} = {n+1 \choose i+1}$ show for n $\geq i \geq 1 : \sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$
i show this with induction:
for n=i=1: ${1+1} \choose {1+1}$ = $2 \choose 2$ = 1 = $0 \choose 1$ + $1 \choose 1$ = $\sum_{k=0}^1 {k \choose 1}$
now let $\sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$
for n+1: ${n+1+1} \choose {i+1}$ = $n+1 \choose i$ + $n+1 \choose i$ = $ n+1 \choose i$ + $\sum_{k=0}^n {k \choose i} = \sum_{k=0}^{n+1} {k \choose i}  $ 
Is this the right way ? 

We can also use the recurrence from Pascal's Triangle and telescoping series:
$$
\begin{align}
\sum_{k=0}^n\binom{k}{i}
&=\sum_{k=0}^n\left[\binom{k+1}{i+1}-\binom{k}{i+1}\right]\\
&=\sum_{k=1}^{n+1}\binom{k}{i+1}-\sum_{k=0}^n\binom{k}{i+1}\\
&=\binom{n+1}{i+1}-\binom{0}{i+1}\\
&=\binom{n+1}{i+1}
\end{align}
$$
