Let $n$ be a nonnegative integer. I am interested in finding a closed form for the sum $$S(x)=\sum_{j=n}^\infty(n+j)(n+j-1)\cdots(j+1)x^j.$$ If I'm not mistaken, we get $$\sum_{j=0}^\infty(n+j)(n+j-1)\cdots(j+1)x^j=\frac{n!}{(1-x)^{n+1}}$$ by differentiating $$\sum_{j=0}^\infty x^j=\frac{1}{1-x}$$ $n$ times.
But what about $S(x)$?
Newton's generalized binomial theorem yields $$\sum_{j=0}^\infty \binom{n+j}{j} x^j = \frac{1}{(1-x)^{n+1}}$$ Your sum is $$S(x)=\frac{1}{n!}\sum_{j=n}^\infty \binom{n+j}{j} x^j = \frac{1}{n!}\left(\frac{1}{(1-x)^{n+1}} - \sum_{j=0}^{n-1} \binom{n+j}{j} x^j\right)$$