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Row $n$ of Pascal's triangle consists of the binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{0}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)$, beginning with row $0$. Boundary entries are one, and Pascal's identity says each interior entry is the sum of the two entries above it. The hockey-stick identity evaluates sums along its diagonals.

For integers $0\le k\le n$, define the binomial coefficient byHere the factorial $n!$ is the product of the positive integers up to $n$. This also counts the $k$-element subsets of an $n$-element set: an ordered choice has $n!/(n-k)!$ possibilities, and each subset is ordered in $k!$ ways.

Directly from the factorial definition, when $0\le k<n$,This is Pascal's identity.

The required hockey-stick identity follows by mathematical induction on $m$, with $n\ge 0$ fixed. For $m=0$, both sides are $1$. If it holds at $m$, adding the next binomial coefficient and using Pascal's identity givesTherefore for every $n,m\ge 0$,