Why does $\lim_{n \to \infty} \sqrt[n]{(-1)^n \cdot n^2 + 1} = 1$? As the title suggests, I want to know as to why the following function converges to 1 for $n \to \infty$:
$$
\lim_{n \to \infty} \sqrt[n]{(-1)^n \cdot n^2 + 1} = 1
$$
For even $n$'s only $n^2+1$ has to be shown, which I did in the following way:
$$\sqrt[n]{n^2} \le \sqrt[n]{n^2 + 1} \le \sqrt[n]{n^3}$$
Assuming we have already proven that $\lim_{n \to \infty}\sqrt[n]{n^k} = 1$ we can conclude that
$$1 \le \sqrt[n]{n^2+1} \le 1 \Rightarrow \lim_{n \to \infty} \sqrt[n]{n^2+1} = 1.$$
For odd $n$'s I can't find the solution. I tried going the same route as for even $n$'s:
$$\sqrt[n]{-n^2} \le \sqrt[n]{-n^2 + 1} \le \sqrt[n]{-n^3}$$
And it seems that it comes down to
$$\lim_{n \to \infty} \sqrt[n]{-n^k}$$
I checked the limit using both Wolfram Alpha and a CAS and it converges to 1. Why is that?

take out the (-1)^n outside the bracket and solve to get -1 as the answer, but didn't get 1
