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?

It's common for CAS's like Wolfram Alpha to take $n$th roots that are complex numbers with the smallest angle measured counterclockwise from the positive real axis. So the $n$th root of negative real numbers winds up being in the first quadrant of the complex plane. As $n\to\infty$, this $n$th root would get closer to the real axis and explain why WA says the limit is 1. CAS's do this for continuity reasons; so that $\sqrt[n]{-2}$ will be close to $\sqrt[n]{-2+\varepsilon\,i}$.
Instead of $\sqrt[n]{x}$, you can get around the issue with $\operatorname{sg}(x)\cdot\sqrt[n]{|x|}$ where $\operatorname{sg}(x)$ is the signum function: $1$ for positive $x$ and $-1$ for negative $x$.
