To what value does this summation converge: $\sum_{r=o}^{n}{\frac{\binom{n-1}{r}r!}{n^{r+1}}}$ I had been trying to solve (Probability of rolling a 1 before you roll two 2's, three 3's, etc) this problem for quite some while and I think I had found some way ahead but I cant seem to find the closed from for the summation that I ended up on. The summation is:
$$\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}$$
I also couldn't come up with worthy bounds for which the summation converges to a particular value(although it does converge as discussed in the original post) by the use of squeeze theorem.
Also WolframAlpha gives the sum as 
$$\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}=\left({\dfrac{e}{n}}\right)^n\Gamma(n,n)$$
Edit:-
I dont know how to handle the $\Gamma(n,n)$ and that is what I need help in which I forgot to mention and it also was the whole point of the post.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{r = 0}^{n - 1}{{n - 1 \choose r}r! \over n^{r + 1}} & =
\sum_{r = 0}^{n - 1}{n - 1 \choose r}\
\overbrace{\int_{0}^{\infty}t^{r}\expo{-nt}\,\dd t}^{\ds{r! \over n^{r + 1}}} =
\int_{0}^{\infty}\expo{-nt}\sum_{r = 0}^{n - 1}{n - 1 \choose r}t^{r}\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\expo{-nt}\pars{1 + t}^{n - 1}\,\dd t =
\expo{n}\int_{1}^{\infty}t^{n - 1}\expo{-nt}\,\dd t =
\expo{n}n^{-n}\
\overbrace{\int_{n}^{\infty}t^{n - 1}\expo{-t}\,\dd t}^{\ds{\Gamma\pars{n,n}}}
\\[5mm] & =
\bbx{\pars{\expo{} \over n}^{n}\,\Gamma\pars{n,n}}
\end{align}

The two arguments $\ds{\Gamma}$ is the Incomplete Gamma Function.

A: As I wrote in comments
$$S_n=\frac{1}{n}\sum_{r=o}^{n-1}{\frac{\binom{n-1}{r}r!}{n^{r}}}=\left(\frac e n \right)^n \,\Gamma(n,n)$$ If you have a look here (formula $8.11.12$), you will find a series expansion for $\Gamma(n,n)$. Using it, you should get 
$$S_n=\sqrt{\frac{\pi }{2}} {\frac{1}{ n^{1/2}}}-\frac{1}{3 n}+ \sqrt{\frac{\pi
   }{2}} \frac{1}{12n^{3/2}}-\frac{4}{135 n^2}+
   \sqrt{\frac{\pi }{2}} \frac{1}{288n^{5/2}}+\frac{8}{2835
   n^3}+O\left(\frac{1}{n^{7/2}}\right)$$ which seems to be very accurate even from small values of $n$ as shown in the table below
$$\left(
\begin{array}{ccc}
n & \text{exact} & \text{approximation} \\
 1 & 1.00000 & 1.00197 \\
 2 & 0.75000 & 0.75020 \\
 3 & 0.62963 & 0.62968 \\
 4 & 0.55469 & 0.55471 \\
 5 & 0.50208 & 0.50209 \\
 6 & 0.46245 & 0.46245 \\
 7 & 0.43116 & 0.43117 \\
 8 & 0.40563 & 0.40563 \\
 9 & 0.38426 & 0.38426 \\
 10 & 0.36602 & 0.36602
\end{array}
\right)$$
