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General Yes, its a L'Hospital's Rule's Application since its an indeterminate power ..
Personally, I prefer substitute $\displaystyle x=\sqrt{n}$ to make the differentiation step easier ..
If $\displaystyle x=\sqrt{n}$, clearly $\displaystyle x\to\infty$ as $\displaystyle n\to\infty$
so $\displaystyle \lim_{n\to\infty} (\sqrt{n}-1)^{\frac{1}{\sqrt{n}}}=\lim_{x\to\infty} (x-1)^{\frac{1}{x}}$ ..
You can evaluate the last one by the L'Hospital's Rule method for the indeterminate powers..
Or you can substitute $\displaystyle t=\frac{1}{x}$ to get a well-known limit ..