Find $\lim\limits_{n\to\infty}{\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) }$ for $k=1, 2, 3, \cdots$ First of all, I already searched Google, math.stackexchange.com...
I know 
$$ \lim_{n\rightarrow\infty} \left( 1+ \frac{1}{n}      \right)      ^n=e$$
That is
$$ \lim_{n\rightarrow\infty} \underbrace{\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\cdots\left(1+\frac{1}{n}\right) }_{\text{n times}}     =e$$
$$$$
At this time, I made some problems modifying above.


*

*$$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\left(1+\frac{n}{n}\right) } =f(1)   $$

*$$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right) }    =f(2)$$

*$$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^3}\right)\left(1+\frac{2}{n^3}\right)\cdots\left(1+\frac{n}{n^3}\right) }    =f(3)$$

*$$ \lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) }    =f(k)$$
$$$$
After thinking above, I feel I'm spinning my wheels with these limit problems.
Eventually, I searched wolframalpha. And the next images are results of wolfram.
(I take a LOG, because I don't know COMMAND of n-times product.)
$$$$




These result (if we trust wolframalpha) say
$$f(1)=\infty$$
$$f(2)=\sqrt{e}$$
$$f(3)=1$$
$$f(30)=1$$
NOW, I'm asking you for help.
I'd like to know how can I find $f(k)$ (for $k=1,2,3,4, \cdots$ ).
I already used Riemann sum, taking Log...  but I didn't get anyhing. ;-(
Thank you for your attention to this matter.
----------- EDIT ---------------------------------
The result for $f(1), f(2), f(3), f(30)$ is an achievement of Wolframalpha, not me.
I'm still spinning my wheel, $f(1), f(2), f(3)$, and so on...
 A: Hint. You may start with
$$
x-\frac{x^2}2\leq\log(1+x)\leq x, \quad x\in [0,1],
$$ giving, for $n\geq1$,
$$
\frac{p}{n^k}-\frac{p^2}{2n^{2k}}\leq\log\left(1+\frac{p}{n^k}\right)\leq \frac{p}{n^k}, \quad 0\leq p\leq n,
$$ and
$$
\sum_{p=1}^n\frac{p}{n^k}-\sum_{p=1}^n\frac{p^2}{2n^{2k}}\leq \sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right)\leq \sum_{p=1}^n\frac{p}{n^k}, \quad 0\leq p\leq n,
$$ or
$$
\frac{n(n+1)}{2n^k}-\frac{n(n+1)(2n+1)}{6n^{2k}}\leq \sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right)\leq \frac{n(n+1)}{2n^k}
$$ and, for $k\geq3$, as $n \to \infty$,
$$
\sum_{p=1}^n\log\left(1+\frac{p}{n^k}\right) \to 0.
$$ that is 

$$
\lim_{n\rightarrow\infty} {\left(1+\frac{1}{n^k}\right)\left(1+\frac{2}{n^k}\right)\cdots\left(1+\frac{n}{n^k}\right) }=1, \quad k\geq3.
$$ 

The cases $k=1, 2$ are clear.
A: Edit: By Taylor series with Lagrange remainder
$$
\ln(1+x)=x-\frac{x^2}{2(1+\xi)^2}
$$
where $0<\xi<x$. So we get 
$$
x-\frac{x^2}{2}\leqslant \ln(1+x)\leqslant x
$$
$$
\frac{n+1}{2}-\frac{(n+1)(2n+1)}{12n}=\sum_{i=1}^{n}\left(\frac{i}{n}-\frac{i^2}{2n^2}\right)\leqslant\sum_{i=1}^{n}\ln(1+\frac{i}{n})\leqslant\sum_{i=1}^{n}\left(\frac{i}{n}\right) =\frac{n+1}{2} 
$$
Thus
$$
\lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n})=\infty
$$
And
$$
\frac{n+1}{2n}-O\left(\frac{1}{n}\right)=\sum_{i=1}^{n}\left(\frac{i}{n^2}-\frac{i^2}{2n^4}\right)\leqslant \sum_{i=1}^{n}\ln(1+\frac{i}{n^2})\leqslant \sum_{i=1}^{n}\left(\frac{i}{n^2}\right)=\frac{n+1}{2n}
$$
Hence
$$
\lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n^2})=\frac1{2}
$$
For any $k>2$, since
$$
O\left(\frac1{n^{k-2}}\right)-O\left(\frac{1}{n^{2k-3}}\right)=\sum_{i=1}^{n}\left(\frac{i}{n^k}-\frac{i^2}{2n^{2k}}\right)\leqslant \sum_{i=1}^{n}\ln(1+\frac{i}{n^k})\leqslant \sum_{i=1}^{n}\left(\frac{i}{n^k}\right)=O\left(\frac1{n^{k-2}}\right)
$$
There is
$$
\lim_{n\to\infty}\sum_{i=1}^{n}\ln(1+\frac{i}{n^k})=0
$$
Note:
This conclusion holds for $k\in\Bbb{R}$, not only $k\in\Bbb{N}$.
