I stuck at the following problem:
Let \begin{equation} Q(n) := \sum_{k \geq 0}\frac{(n-1)_k}{n^k} \end{equation} where $(n)_k = n (n-1) \ldots (n-k + 1)$. I want to show the following equation: \begin{equation} 1 + Q(n) = \int^{\infty}_{0}{e^{-x}\left(1 + \frac{x}{n} \right)^n dx} \end{equation} By induction for $n \geq 1$. For $n = 1$ this is clear. For the induction step I got \begin{equation} Q(n+1) = \sum_{k \geq 0}{\frac{(n)_k}{(n+1)^k}} = \sum_{k \geq 0}{\frac{(n-1)_k}{(n+1)^k}\frac{n}{n-k}} \end{equation} and \begin{equation} \int^{\infty}_{0}{e^{-x}\left(1 + \frac{x}{n+1} \right)^{n+1} dx} = 1 + \int^{\infty}_{0}{e^{-x} \left( 1 + \frac{x}{n+1}\right)^n dx} \end{equation} I want to connect $Q(n+1)$ and the integral expression above by the induction requirement. I used substitution for the integral but this wasn't helpful. I would be thankful if anyone can give me a hint.
There is no need of induction, it actually increases our effort. Simplification of the integral will directly give us the summation. $$\int_{0}^{\infty}e^{-x}\left(1+\frac{x}{n}\right)^{n}dx$$ Using the binomial theorem we can expand $\left(1+\frac{x}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{x}{n}\right)^{k}$. $$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^{k}}\int_{0}^{\infty} x^{k}e^{-x}dx$$ The expression inside the integral is gamma function or factorial function. $$\sum_{k=0}^{n}\binom{n}{k}\frac{k!}{n^{k}}$$ Note that, $k! \binom{n}{k}=\frac{n!}{(n-k)!}=(n-k+1)(n-k+2)..(n)$. So summation is, $$\sum_{k=0}^{n}\frac{(n)...(n-k+2)(n-k+1)}{n^{k}}$$ Which proves the required claim