Worpitzky identity for multipermutations (Q2435791)

Let \({\mathbf{n}}:=1^{k_1}\ldots n^{k_n}\) be a multiset with \(k_i\) repetitions of \(i\) for \(1\leq i\leq n\). The set \(S_{\mathbf{n}}\) of all permutations of \({\mathbf{n}}\) is the set of all mappings \(\sigma\colon\{1,2,\ldots,k_1+\ldots+k_n\}\to\{1,2,\ldots,n\}\) such that the cardinality \(| \sigma^{-1}(i)|\) equals \(k_i\) for all \(i\). The element \(i\) is called an index of lowering if either \(i=n\) or \(i<n\) and \(\sigma(i)>\sigma(i+1)\). Let \(a_{{\mathbf{n}},p}\) be the number of all \(\sigma\in S_{\mathbf{n}}\) with exactly \(p\) indices of lowering (Euler numbers of \({\mathbf{n}}\)). Then the author shows that for \({\mathbf{n}}:=1^{k_1}\ldots n^{k_n}\) the identity for \(x\) \[ \displaystyle \prod_{i=1}^n \binom{x+k_i-1}{k_i}=\sum_{p>0}\binom{x+k_1+\ldots+k_n-p}{k_1+\ldots+k_n}a_{{\mathbf{n}},p} \] holds true. For the special case \(k_1=k_2=\ldots=k_n=1\) using \(a_{n,p}:=a_{1^1\ldots n^1,p}\) this gives a well-known identity for these (ordinary) Euler numbers.