Finding $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ $\large\zeta_7\left(\zeta_3\right)^5$ where $\large\zeta_n=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$
I am having trouble getting a final answer that makes sense to me. Here is what I tried:
$\large\zeta_7\left(\zeta_3\right)^5=\left(\large\zeta_{21}\right)^{3}\left(\zeta_{21}\right)^{35}=\left(\zeta_{21}\right)^{38}=\cos{\frac{76\pi}{21}}+i\sin{\frac{76\pi}{21}}$
Is this an acceptable method/answer? Intuitively, it doesn't feel right.
A second attempt yielded:
$\large\zeta_7=\cos{\frac{2\pi}{7}}+i\sin{\frac{2\pi}{7}}$
$\left(\large\zeta_3\right)^5=\cos{\frac{10\pi}{3}}+i\sin{\frac{10\pi}{3}}$
From here, converting them to the form: $\large e^{\left(\frac{2\pi i}{n}\right)}$
$\large\zeta_7=\large e^{\left(\frac{2\pi i}{7}\right)}$
$\left(\large\zeta_3\right)^5=\large e^{5{\left(\frac{2\pi i}{3}\right)}}$
Multiplying these together: $\large e^{\left(\frac{2\pi i}{7}\right)}\large e^{5{\left(\frac{2\pi i}{3}\right)}}=\large e^{\left(\frac{76\pi i}{21}\right)}$. 
This gives an equivalent answer to above which means I am either right, or being tricked into thinking I'm right by my poor math. Any help would be greatly appreciated.

Euler's formula makes this straightforward:  $$\zeta_n = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n} = e^{2\pi i/n}.$$   Consequently,  $$\zeta_7 \zeta_3^5 = e^{2 \pi i/7} (e^{2 \pi i/3})^5 = e^{2\pi i/7 + 10 \pi i/3} = e^{76\pi i/21} = e^{34 \pi i/21} e^{2\pi i} = \cos \frac{34\pi}{21} + i \sin \frac{34\pi}{21},$$ where we used the fact that $e^{2\pi i} = e^0 = 1.$
