Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form

$$ \color{red}{\mathbf 5}\cdot\sum_{k\geqslant0}{n\choose \color{red}{\mathbf 5}k}=\sum_{\ell=1}^{\color{red}{\mathbf 5}}\left(1+\mathrm e^{2\mathrm i\pi\ell/\color{red}{\mathbf 5}}\right)^n=\sum_{\zeta\in\mathbb C\,:\,\zeta^\color{red}{\mathbf 5}=1}\left(1+\zeta\right)^n $$

Hint: with $\omega=\exp(2\pi i /5)$ a primitive $5$th root of unity, we have

$$\sum_{r=0}^4 \omega^{rk}=\begin{cases}5 & k\equiv 0 \bmod 5 \\ 0 & k\not\equiv 0\bmod 5.\end{cases}$$

So then what is

$$(1+\omega^0)^n+(1+\omega^1)^n+(1+\omega^2)^n+(1+\omega^3)^n+(1+\omega^4)^n~? $$

(Combine the binomial expansions...)

As you know, $\sum_{j=0}^n {n \choose j} = 2^n$, and if $\omega$ is a primitive $5$'th root of unity $\sum_{j=0}^n {n \choose j} \omega^j = (1 + \omega)^n$. Now $\sum_{i=0}^4 \omega^{ij} = 5$ if $j$ is divisible by $5$ and $0$ otherwise. So $$\sum_{k} {n \choose {5k}} = \frac{1}{5} \sum_{i=0}^4 \sum_{j=0}^n {n \choose j} \omega^{ij} = \frac{1}{5} \sum_{i=0}^4 (1+\omega^i)^n$$