$t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$. $t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$.
I wanted to make a proof by induction. I've checked it for $n = 3$:
\begin{equation}
\begin{split}
\prod^{3-1}_{k=0}(t-C^k) & = (t-1)(t-e^{\frac{2\pi i}{3}})(t-e^{\frac{4\pi i}{3}}) \\
 & = (t-1)(t-\cos(\frac{2}{3}\pi)+i\sin(\frac{2}{3}\pi))(t-\cos(\frac{4}{3}\pi)+i\sin(\frac{4}{3}\pi)) \\
 & = (t-1)(t+\frac{1}{2}+\frac{\sqrt{3}}{2}i)(t+\frac{1}{2}-\frac{\sqrt{3}}{2}i) \\
 & = (t-1)(t^2+\frac{1}{2}t-\frac{\sqrt{3}}{2}it+\frac{1}{2}t+\frac{1}{4}-\frac{\sqrt{3}}{4}i+\frac{\sqrt{3}}{2}it+\frac{\sqrt{3}}{4}i+\frac{3}{4}) \\
 & = (t-1)(t^2+t+1) = (t^3-1)
\end{split}
\end{equation}
This gave me an idea because
\begin{equation}
\prod^{n-1}_{k=0}(t-C^k) = (t-1)\prod^{n-1}_{k=1}(t-C^k)
\end{equation}
To show that
\begin{equation}
\prod^{n}_{k=1}(t-C^k) = \sum^{n}_{k=0}t^k
\end{equation}
But I can't seem to figure out how. What also makes this proof difficult is that if $n$ increases $C$ changes also. Can anybody help me?

You are overcomplicating the question. Just note that the roots of $t^n-1$ are exactly $C^k$ with $k$ between $0$ and $n-1$. Just factorice the polynomial and get the result.
