Combinatorial proof of $\sum_{k=0}^n k \cdot k! = (n+1)! -1$

Solution 1:

$k\cdot k!$ is the number of permutations of the $(n+1)$ symbols $0,\,1,\,\dotsc,\, n$ such that $k$ is the largest symbol that is not kept fixed. (Symbol $k$ can be mapped to the $k$ places $0,\,\dotsc,\,k-1$, the $k$ smaller symbols then can be placed arbitrarily in the $k$ left free places.) So

$$\sum_{k=0}^n k\cdot k!$$

is the number of non-identity permutations of the $(n+1)$ symbols. On the other hand, that number is of course the total number of permutations of $(n+1)$ symbols minus one.

Not the best combinatorial proof ever, but meh.

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