Evaluate $(1+p)\cdots(1+\frac pn)$ 
Prove for $\forall p\in\mathbb{R}$, 
  $$\lim_{n\rightarrow\infty}\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}=L_p$$
  where $L_p\in\mathbb{R}$.

Moreover, I tried to calculate $L_p$, and it can be calculated if $p$ is an integer. What if $p$ is not an integer? 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
L_{p} & \equiv \lim_{n \to \infty}{\prod_{k = 1}^{n}\pars{1 + p/k} \over n^{p}} =
\lim_{n \to \infty}\bracks{{1 \over n^{p}}\,
{\prod_{k = 1}^{n}\pars{k + p} \over n!}} =
\lim_{n \to \infty}\bracks{{1 \over n^{p}n!}\,\pars{1 + p}^{\large\overline{n}}}
\\[5mm] & =
\lim_{n \to \infty}\bracks{{1 \over n^{p}n!}\,{\Gamma\pars{1 + p + n} \over \Gamma\pars{1 + p}}} =
{1 \over \Gamma\pars{1 + p}}\lim_{n \to \infty}
\bracks{{1 \over n^{p}}\,{\pars{n + p}! \over n!}}
\\[5mm] & =
{1 \over \Gamma\pars{1 + p}}\,\lim_{n \to \infty}\bracks{{1 \over n^{p}}\,
{\root{2\pi}\pars{p + n}^{p + n + 1/2}\expo{-p - n} \over \root{2\pi}n^{n + 1/2}\expo{-n}}}\quad\pars{\substack{Here,\ I\ use\ the\ well\ known\\[1mm] {\large z!\ Stirling\ Asymptotic\ Expansion} }}
\\[5mm] & =
{1 \over \Gamma\pars{1 + p}}\,\lim_{n \to \infty}\bracks{{1 \over n^{p}}\,
{n^{n + p + 1/2}\,\pars{1 + p/n}^{3/2 + p + n}\,\expo{-p} \over n^{n + 1/2}}}
\\[5mm] & =
{1 \over \Gamma\pars{1 + p}}\,\lim_{n \to \infty}
\bracks{\pars{1 + {p \over n}}^{n}\,\expo{-p}} = \bbx{1 \over \Gamma\pars{1 + p}}
\end{align}
A: $$\lim_{n\rightarrow\infty}\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}=1/\Gamma(1+p)$$
Let me know if you need hints.
I am sure I am the first who got this correct.
A: Since the result is
$\dfrac1{\Gamma(p+1)}$,
I'll try to get
$\dfrac1{p!}$
for integer $p$
in an relatively elementary way.
$\begin{array}\\
L(p, n)
&=\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}\\
&=\dfrac{\prod_{k=1}^n(\frac{k+p}{k})}{n^p}\\
&=\dfrac{\prod_{k=1}^n(k+p)}{n^p\prod_{k=1}^nk}\\
&=\dfrac{\prod_{k=p+1}^{n+p}k}{n^p\prod_{k=1}^nk}\\
&=\dfrac{\prod_{k=p+1}^{n}k\prod_{k=n+1}^{n+p}k}{n^p\prod_{k=1}^pk\prod_{k=p+1}^nk}
\qquad\text{assuming } n > p\\
&=\dfrac{\prod_{k=n+1}^{n+p}k}{n^p\prod_{k=1}^pk}\\
&=\dfrac{\prod_{k=1}^{p}(n+k)}{n^pp!}\\
&=\dfrac{\prod_{k=1}^{p}(1+k/n)}{p!}\\
&>\dfrac{1}{p!}\\
\text{and}\\
L(p, n)
&<\dfrac{(1+p/n)^p}{p!}\\
&<\dfrac{(1+1/n^{2/3})^{n^{1/3}}}{p!}
\qquad\text{if } n > p^3\\
&<\dfrac{e^{1/n^{1/3}}}{p!}
\qquad\text{since } (1+1/m^2)^m < e^{1/m}\\
&\to\dfrac{1}{p!}
\qquad\text{as } n \to \infty\\
\end{array}
$
A: This is not an answer but it is too long for a comment.
Using Felix Marin's elegant answer and  approach, we could even approximate the value of the partial terms
$$P_{n,p}=\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}$$ extending the  series expansion  of
$$A=\frac 1 {n^p}\frac{ (n+p)!}{n!}\implies \log(A)=   \log \left(\frac 1 {n^p}\frac{ (n+p)!}{n!}\right)$$
using Stirling approximation; when this is done, Taylor again using $A=e^{\log(A)}$ leads to $$P_{n,p}=\frac{1}{\Gamma (1+p)} \left(1+\frac{p (p+1)}{2 n}+ \frac{(3p+2)(p+1)p(p-1)}{24n^2 }\right)+O\left(\frac{1}{n^3}\right)$$ which appears to be quite good even for rather small values of $n$
$$\left(
\begin{array}{cccc}
 p & n & \text{exact} & \text{approximation} \\
 \frac{1}{\pi } & 5 &  1.16285 & 1.16281 \\
 \frac{1}{\pi } & 10 & 1.14055 & 1.14054 \\
 \frac{1}{\pi } & 15 & 1.13295 & 1.13294 \\
 \frac{1}{\pi } & 20 & 1.12911 & 1.12911 \\
 & & & \\
 \pi  & 5  & 0.40364 & 0.39394 \\
 \pi  & 10 & 0.24928 & 0.24808 \\
 \pi  & 15 & 0.20801 & 0.20766 \\
 \pi  & 20 & 0.18914 & 0.18899
\end{array}
\right)$$
