If \( P_{m} \) stands for \( { }^{m} P_{m} \), then prove that: \[ 1+1 \cdot P_{1}+2 \cdot P_{2}...
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0If \( P_{m} \) stands for \( { }^{m} P_{m} \), then prove that:
\[
1+1 \cdot P_{1}+2 \cdot P_{2}+3 \cdot P_{3}+\cdots+n \cdot P_{n}=(n+1) !
\]
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