Here is a completely elementary proof, which only needs introductory calculus concepts:
$$a_{n+1} = \frac{(n+1)!}{(a+1)(a+2)\dots(a+n+1)} = \frac{1}{a-1}\left(\frac{(n+1)!}{(a+1)(a+2) \dots (a+n)} -\frac{(n+2)!}{(a+1)(a+2) \dots (a+n+1)}\right) = b_{n+1} - b_{n+2}$$
where $a_n$ is the term of our series and $$b_n = \frac{1}{a-1}\left(\frac{n!}{(a+1)(a+2) \dots (a+n-1)}\right)$$ with $b_1 = \frac{1}{a-1}$
Thus the sum we seek, telescopes!
Giving us
$$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (b_{k} - b_{k+1}) = b_1 - b_{n+1}$$
Thus we just need to compute $\lim b_n$
(we basically need to show that $b_n \to 0$ to match the limit being $\frac{1}{a-1}$)
Now we have that $a \gt 1$, so let $a = 1 + x$ for $x \gt 0$.
$$ (a-1)b_n =\frac{n!}{(a+1)(a+2) \dots (a+n-1)} = \frac{1}{(1 + x/2)(1+x/3)\dots(1+x/n)} $$
Let $ M = \lceil x \rceil$ and consider the product
$$p_n = \prod_{k=M}^{n} \left(1 + \frac{x}{k}\right)$$
It is enough to show that $\log p_n \to \infty$ (that proves that $b_n \to 0$).
Now we have that $\dfrac{1}{1+t} \gt 1-t$ for $0 \lt t \le 1$ and thus integrating between $0$ and $y$ (where $y \le 1$) we get that
$$ \log(1+y) \ge y - \dfrac{y^2}{2}$$
Now $$\log p_n = \sum_{k=M}^{n} \log (1 + \frac{x}{k})$$
$$ \ge \sum_{k=M}^{n} (\frac{x}{k} - \frac{x^2}{2k^2})$$
This goes to $\infty$ as the harmonic series diverges, and the sum of reciprocals of the squares converges.
Thus the sum of your sequence is $$b_1 = \frac{1}{a-1}$$