If you want to use induction, note that, for any real (or complex) numbers $\phi$ and $\theta$, we have
$$2\,\cos(\theta)\,\cos(\phi)=\cos(\phi+\theta)+\cos(\phi-\theta)\,.\tag{*}$$
Now, let $$f_n(\theta):=\sum_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)$$ for each integer $n\geq 0$ and for each real (or complex) number $\theta$. We claim that $f_n(\theta)=2^n\,\cos^n(\theta)$. The basis step $n=0$ is trivial.
Now, for the inductive step, let $n\in\mathbb{Z}_{\geq 0}$ be such that $f_n(\theta)=2^n\,\cos^n(\theta)$. That is,
$$2^n\,\cos(\theta)=\sum\limits_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)\,.$$
Therefore,
$$2^{n+1}\,\cos^{n+1}(\theta)=2\,\cos(\theta)\,\big(2^n\,\cos^n(\theta)\big)=2\,\cos(\theta)\,\left(\sum\limits_{k=0}^n\,\binom{n}{k}\,\cos\big((n-2k)\theta\big)\right)\,.$$
Therefore,
$$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^n\,\binom{n}{k}\,2\,\cos(\theta)\,\cos\big((n-2k)\theta\big)\,.$$
From (*), we see that
$$\begin{align}2\,\cos(\theta)\,\cos\big((n-2k)\theta\big)&=\cos\big((n-2k)\theta+\theta\big)+\cos\big((n-2k)\theta-\theta\big)
\\
&=\cos\Big(\big((n+1)-2k\big)\theta\Big)+\cos\Big(\big((n+1)-2(k+1)\big)\theta\Big)\,.\end{align}$$
This shows that
$$\begin{align}2^{n+1}\,\cos^{n+1}(\theta)&=\sum_{k=0}^n\,\binom{n}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)+\sum_{k=0}^n\,\binom{n}{k}\,\cos\Big(\big((n+1)-2(k+1)\big)\theta\Big)
\\
&=\sum_{k=0}^{n+1}\,\binom{n}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)+\sum_{k=0}^{n+1}\,\binom{n}{k-1}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)\,,
\end{align}$$
where $\displaystyle\binom{n}{n+1}$ and $\displaystyle\binom{n}{-1}$ are conventionally defined to be $0$. Ergo,
$$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^{n+1}\,\Biggl({{\binom{n}{k}+\binom{n}{k-1}}}\Biggr)\,\cos\Big(\big((n+1)-2k\big)\theta\Big)\,.$$
By Pascal's Rule,
$$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$$
for all $k=0,1,2,\ldots,n+1$. It follows immediately that
$$2^{n+1}\,\cos^{n+1}(\theta)=\sum_{k=0}^{n+1}\,\binom{n+1}{k}\,\cos\Big(\big((n+1)-2k\big)\theta\Big)=f_{n+1}(\theta)\,.$$
The proof is now complete.