proof of a combinatorial identity using inclusion-exclusion principle: $ \sum _{k=m} ^{n} (-1)^{k-m} {n \choose k} = {n-1 \choose m-1}$

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How to prove the following using inclusion exclusion

$$ \sum _{k=m} ^{n} (-1)^{k-m} {n \choose k} = {n-1 \choose m-1}$$