General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)…(k+m)}$

Is there a general formula for

$\sum_{k=1}^n\frac{1}{k(k+1)…(k+m)}$?

I know that the limit is $\frac{1}{mm!}$ but is there a combinatorial expression for this?

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