The character table of a finite cyclic group over $\mathbb{C}$

By [F1] the entry in row ${\chi }_k$ and column $g^m$ is ${\zeta }^{km}$, so the table has the stated shape.

If $n=1$, then $G$ is trivial and ${\sum }_{m=0}^{n-1}{\zeta }^{km}=1$ for the unique value $k=0$. Assume now that $n>1$. Then the polynomial $t^n-1$ has coefficient $0$ at $t^{n-1}$, so [A1] and [F2] give ${\sum }_{m=0}^{n-1}{\zeta }^m=0$. More generally, if $k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then ${\sum }_{m=0}^{n-1}{\zeta }^{km}=n$; if $0<k<n$, the element $\eta ={\zeta }^k$ has order $n\mathrm{/}d$, where $d=\gcd (k,n)$, so ${\sum }_{m=0}^{n-1}{\zeta }^{km}=d{\sum }_{r=0}^{n\mathrm{/}d-1}{\eta }^r\mathrm{.}$ Here $n\mathrm{/}d>1$, and the inner sum vanishes by the same Vieta argument applied to $t^{n\mathrm{/}d}-1$. Thus ${\sum }_{m=0}^{n-1}{\zeta }^{km}$ is $n$ when $k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise.

Row orthogonality: by [F3] and step 1.1, ${\delta }_{kl}=⟨{\chi }_k,{\chi }_l⟩=\frac{1}{n}{\sum }_m{\zeta }^{km}\overline{{\zeta }^{lm}}=\frac{1}{n}{\sum }_m{\zeta }^{(k-l)m}$, and the last sum is $n$ when $k\equiv l(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise by step 1.2, which matches ${\delta }_{kl}$.

Column orthogonality: by [F4] and step 1.1, ${\sum }_k{\chi }_k(g^m)\stackrel{‾}{{\chi }_k(g^{m^{\mathrm{\prime }}})}={\sum }_k{\zeta }^{k(m-m^{\mathrm{\prime }})}$, which is $n$ when $m\equiv m^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise by step 1.2; since $G$ is abelian every centralizer is $G$, so this is exactly $\mathrm{\mid }{C}_G(g)\mathrm{\mid }{\delta }_{m{m}^{\mathrm{\prime }}}$.

All $n$ degrees are $1$, so ${\sum }_k{n}_k^2=n=\mathrm{\mid }G\mathrm{\mid }$.