closed formula for Sum of this series Is there any closed form of this series?

$$\frac{x!}{1!} + \frac{(x+1)!}{2!} + \frac{(x+2)!}{3!} +\cdots + \frac{(x+n-1)!}{n!}$$

I tried to manipulate the expansion of $(1+x)^n$ but can't seem to get the term $(1+x)!$ from $x!$. Need help on this one.

By the Pascal formula we get:
$$\begin{align}
\frac{x!}{1!}+\dots+\frac{(x+n-1)!}{n!}&=(x-1)!\left(\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}\right)\\\\
&=(x-1)!\left(\binom{x}{x}+\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}-1\right)\\\\
&=(x-1)!\left(\binom{x+n}{x}-1\right)\\\\
&=x\cdot\frac{(x+n)!}{n!}-(x-1)!
\end{align}$$
Where I added and subtracted a term $\binom{x}{x}$ so I can use the Pascal formula.
