$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k$ by Stolz–Cesàro $$\lim_{n \to \infty} \frac{1}{\sqrt{n^k}} (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k, k \in \mathbb{N}$$
Putting $\sqrt{n^k}$ in denominator, we get:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k}{\sqrt{n^k}}$$
Apllying Stolz–Cesàro theorem:
$$\lim_{n \to \infty} \frac{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}})^k - (1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n-1}})^k }{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
I don't know how to use properly multinomial formula, but the biggest number in numerator among denominators in first brackets with $k$ degree is $\sqrt{n^k}$, in second brackets $\sqrt{(n-1)^k}$, their coefficients are $1$
$$\lim_{n \to \infty} \frac{ \frac{1}{\sqrt{n^k}} - \frac{1}{\sqrt{(n-1)^k}} + \frac{1}{\sqrt{(n-1)^k}} - \cdots + 1 - 1 + \cdots}{\sqrt{n^k} - \sqrt{(n-1)^k}}$$
Multiplying numbers in first brackets by $\frac{1}{\sqrt{n}}$ gives us $\frac{1}{\sqrt{2^{k-1} \times n}} + \frac{1}{\sqrt{3^{k-1} \times n}} + \cdots + \frac{1}{\sqrt{(n-1)^{k-1} \times n}}$ etc., but I suspect we can't get same numbers in second brackets.
My guess, after evaluating sum (or difference) in numerator, we need to divide each number of our expression by $\sqrt{n^k}$.
In denominator we get 0, because $$\sqrt{\frac{n^k}{n^k}} - \sqrt{\frac{(n-1)^k}{n^k}} = 1 - \sqrt{(\frac{n-1}{n})^k} = 1 - 1 = 0, n \to \infty$$
Multiplication and division of denominator by conjugate ($\sqrt{n^k} + \sqrt{(n-1)^k}$) gives "slightly" different picture (unless I've done several big mistakes in a row):
$$\frac{ n^k - (n-1)^k }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{ n^k - n^k + kn^{k-1} - \cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} } = \frac{kn^{k-1} - {k\choose 2}n^{k-2} +\cdots }{ \sqrt{n^k} + \sqrt{(n-1)^k} }, n \to \infty$$
Dividing by $n^{k-1}$:
$$\frac{k - 0 + 0}{ \sqrt{\frac{n^k}{n^{2(k-1)}  } } + \sqrt{ \frac{(n-1)^k}{n^{2(k-1)}}  }} = \frac{k - 0 + 0 - \cdots}{0 + 0}$$
So my guess was a bad guess.
Unfortunately, by this point I'm pretty much lost. Hence, I need some help... Thank you!

There is a nice proof already provided by iskander using Stolz–Cesàro.
Also, using definite integrals it can be shown the limit is $2^k$
$\left( \lim_{n \to \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\sqrt{i}} \right)^k=\left( \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n}\sqrt{\frac{n}{i}} \right)^k=\left( \int_{0}^{1}\frac{dx}{\sqrt
{x}} \right)^k=2^k$
