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.

For this you need to know that $$(A \text{ cis } B) \cdot (C \text{ cis } D) = (AC) \text{ cis } (B + D)$$
And that 
$$(A \text{ cis } B)^n = \left(A^n \text{ cis } Bn\right)$$
Your problem is
$$\large\zeta_7\left(\zeta_3\right)^5$$
$$\left(1 \text{ cis } \frac{2\pi}{7}\right)\left(1 \text{ cis } \frac{2\pi}{3}\right)^5$$
$$\left(1 \text{ cis } \frac{2\pi}{7}\right)\left(1^5 \text{ cis } \frac{5\cdot 2\pi}{3}\right)$$
$$\left(1 \cdot 1^5 \right)\text{ cis } \left(\frac{2\pi}{7}+ \frac{5\cdot 2\pi}{3}\right)$$
$$1\text{ cis } \left(2\pi\frac{38}{21}\right)$$
And give that radians repeat every $2\pi$:
$$1\text{ cis } \left(2\pi\frac{17}{21}\right)$$
$$\left(1\text{ cis } \frac{2\pi}{21}\right)^{17}$$
$$\left(\zeta_{21}\right)^{17}$$
