Consider the following "more conventional" statement (adjust notation accordingly) of the problem:
Problem: Show that for each $n\geq 0$,
$$
\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}.
$$
Proof: For each $n\geq 0$, let $S(n)$ be the declaration that for every $m\geq 0$,
$$
\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}.
$$
Base step: $S(0)$ says that $\sum_{i=0}^0\binom{m+i}{i}=\binom{m+1}{0}$, which is true because both sides are equal to $1$.
Induction step: For some $k\geq 0$, assume that $S(k)$ is true. To be shown is that $S(k+1)$ is true; that is, for any $m\geq 0$,
$$
\sum_{i=0}^{k+1}\binom{m+i}{i}=\binom{m+k+2}{k+1}.
$$
Beginning with the LHS,
\begin{align}
\sum_{i=0}^{k+1}\binom{m+i}{i} &= \sum_{i=0}^k\binom{m+i}{i}+\binom{m+k+1}{k+1}\tag{defn. of $\sum$}\\[1em]
&= \binom{m+k+1}{k}+\binom{m+k+1}{k+1}\tag{by $S(k)$}\\[1em]
&= \binom{m+k+2}{k+1},\tag{Pascal's identity}
\end{align}
we obtain the RHS.
By mathematical induction, then, for all $n\geq 0, S(n)$ is true. $\Box$