Prove combinatorically $$P_r^{n+1}=r!+r(P_{r-1}^n+P_{r-1}^{n-1}+\cdots+P_{r-1}^r),$$ where $P_r^n$ denotes the number of $r-$ permutations of an $n$ element set.
My attempt:
The LHS denotes the number of $r$ permutations of a $n+1$ element set. $P_r^{n+1}=\binom{n+1}rr!$ so we can view that as the total number of total orderings on $r$ element subsets of an $n+1$ element set.
On the RHS, $r!=r(r-1)!=r\binom{r-1}{r-1}(r-1)!=rP_{r-1}^{r-1}$, so we can rewrite the RHS as: $$\begin{aligned}r!+(P_{r-1}^n+P_{r-1}^{n-1}+\cdots P_{r-1}^r)&=r\sum_{k=r-1}^nP_{r-1}^k\\&=r\sum_{k=r-1}^n \binom{k}{r-1}(r-1)!\\&=r!\sum_{k=r-1}^n\binom{k}{r-1}\end{aligned}.$$ Now, $\binom{k}{r-1}$ can represent the number of ways to choose a subset of $r$ elements of a set $\{1,2,\ldots,n+1\}$ with the greatest element $k+1$. Hence, $\sum_{k=r-1}^n\binom{k}{r-1}=\binom{n+1}r$. So, the RHS is $r!\binom{n+1}k,$ but I don't like converting $P_r^n$ to $\binom{n}rr!.$ Is there any more illustrative and elegant way?
Your proof is rigorous and elegant. Here is another perspective.
$P^{n+1}_r$ is the number of ways to make an ordered selection of $r$ items from $\{1,2,\dots,n+1\}$.
For each $k\in \{r-1,r,\dots,n\}$, on the other hand, $r\cdot P^k_{r-1}$ is the number of such selections where the largest selected element is $k+1$. Indeed, there are $r$ places to place $k+1$, and then $P^k_{r-1}$ ways to fill the other $r-1$ spots with elements from $\{1,\dots,k\}$.
Summing over all such $k$, the identity $P^{n+1}_r=r\sum_{k=r-1}^n P^k_{r-1}$ follows.