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 $k=1, 2, 3, \cdots$

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.