How to prove that $$\sum_{k=1}^n k \frac{(n-1)!}{(n-k)!} n^{n-k} = n^n$$ I have tried induction but changing from $n$ to $(n+1)$ even further complicate the formula and no way to extract the case $n$ for substitution as $n^n$. I also considered the binomial expansion of $(z+n)^n$ and differentiate it to obtain the $k$ in the summation. But I can't eliminate the $1/k!$ at the denominator so as to match the original RHS.
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2This problem looks very similar to me. – Sarvesh Ravichandran Iyer Jul 25 '22 at 04:00
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1Does this answer your question? How to show $\sum_{k=1}^n \binom{n-1}{k-1} n^{n-k} k! = n^n$ - found using Sarvesh's comment above. Note $\binom{n-1}{k-1}k! = \left(\frac{(n-1)!}{(k-1)!(n-k)!}\right)k! = k\frac{(n-1)!}{(n-k)!}$, so it's more than just similar, with it actually just being a version written somewhat differently. – John Omielan Jul 25 '22 at 05:45
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Yes, the two are the same. Thank you for the reference. – LM Cheong Jul 25 '22 at 07:32