Solution $1:$ note $(z-e^{i\alpha})(z-e^{-i\alpha})=z^2-2z\cos\alpha+1$ hence:
$$\big(z^{n}-1\big)^2=\prod_{k=1}^n \Big( z^2-2z\cos
\left(\frac{2k}{n}\pi\right)+1\Big)\quad (\star)$$
Writing out the hint:
$$\lim_{n\to\infty} \sqrt[n]{\prod_{k=1}^n \Big( z^2-2z\cos \left(\frac{2k}{n}\pi\right)+1\Big)}=\exp\left\{\frac{1}{2\pi}\int_0^{2\pi}\ln \big(z^2-2z\cos t+1\big)\,\mathrm{d}t\right\}$$
Given $(\star)$ and $n\to\infty:\;\sqrt[n]{\big(z^{n}-1\big)^2}\to z^2$, the result follows.
Alternative approaches:
Solution $2:$ $\text{Log}\,z$ is holomorphic on $\mathbb{C}\setminus \mathbb{R}_{\leq 0}$, in particular on $\{z : |z −r| =1,\;r\in\mathbb{R}_{>1}\}$. Tweeking Gauss' MVT:
$$\ln r=\frac{1}{2\pi}\int_0^{2\pi}\text{Log} (r-e^{it})\,\mathrm{d}t\quad\text{and}\quad \ln r=\frac{1}{2\pi}\int_0^{2\pi}\text{Log} (r-e^{-it})\,\mathrm{d}t$$
Adding both yields the result.
Solution $3:$ as requested, below is a sketch for a solution which presumes no knowledge of $\mathbb{C}.$
$$\text{Define for}\;\lambda\geq 1:\;f(\lambda)=\int_0^{2\pi}\ln \big(\lambda^2-2\lambda\cos x+1\big)\,\mathrm{d}x\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}\lambda}=\int_0^{2\pi}\cdots=\frac{4\pi}{\lambda}$$
The latter can be done using Weierstraß and is straightforward. Furthermore,
$$\begin{aligned}\frac{1}{2}f(1)-\pi\ln 2 &=\int_0^{\pi}\ln (1\pm\cos x)\mathrm{d}x\\&=\underbrace{\int_0^{\pi}\ln \sin x\,\mathrm{d}x}_{=\,-\pi\ln 2\;(\clubsuit)}\Rightarrow f(1)=0\end{aligned}\;\;\Longrightarrow\;\; \frac{1}{2\pi}f(r)=\int_1^{r}\frac{2\,\mathrm{d}\lambda}{\lambda}=2\ln r$$
$($the claimed result in $(\clubsuit)$ is a classic$)$
