Generalized Eulerian polynomials

Define polynomials $$S_{k,n}(x)$$ by $$\sum_{j\ge 0}\binom{k+j}{k}^n x^j=\frac{ S_{k,n}(x)}{(1-x)^{k n+1}},$$ which for $$k=1$$ reduce to the Eulerian polynomials.

Computatios suggest that $$S_{k,n}(1)=\frac{(kn)!}{k!^n}.$$

Any idea how to prove this? Have these polynomials been studied in the literature?

Suppose that $$a(x)=\sum\limits_{j\geqslant 0}a_j x^j$$ and $$b(x)=\sum\limits_{j\geqslant 0}b_j x^j$$ converge when $$|x|<1$$, with $$b_j>0$$, and $$\lim\limits_{x\to 1^-}b(x)=\infty$$. If $$\lim\limits_{j\to\infty}\dfrac{a_j}{b_j}=\lambda$$ exists then $$\lim\limits_{x\to 1^-}\dfrac{a(x)}{b(x)}=\lambda$$.
Take $$a_j=\binom{k+j}{k}^n$$, $$b_j=\binom{kn+j}{kn}$$, and use $$\lim\limits_{j\to\infty}\frac1{j^m}\binom{m+j}{m}=\frac1{m!}$$, $$\sum\limits_{j\geqslant 0}\binom{m+j}{m}x^j=(1-x)^{-m-1}$$.