= Solution
Fix $m\geq0$, using the convention for <binomial coefficients> that $\binom{k}{m}=0$ when $m>k$. At $n=0$, both $\binom0m$ and $\binom1{m+1}$ equal $1$ if $m=0$ and $0$ otherwise.
For the successor step, apply the <induction hypothesis>, then <Pascal's rule>:
$$
\sum_{k=0}^{n+1}\binom{k}{m}
=\binom{n+1}{m+1}+\binom{n+1}{m}
=\binom{n+2}{m+1}.
$$
Thus <mathematical induction> proves the <hockey-stick identity>
$$
\boxed{\sum_{k=0}^{n}\binom{k}{m}=\binom{n+1}{m+1}\quad(m,n\geq0)}.
$$
The zero convention also covers the case $n<m$, without a separate starting index.
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