Bessel's inequality

Decomposition. Let $p=\sum_i\langle v,e_i\rangle e_i$. Then $\langle p,e_j\rangle=\langle v,e_j\rangle$ for each $j$, so $v-p\perp e_j$ for all $j$.

Norm expansion.
$$\|v\|^2=\|p+(v-p)\|^2=\|p\|^2+\|v-p\|^2$$
because $p\perp(v-p)$.

Computation. $\|p\|^2=\sum_i|\langle v,e_i\rangle|^2$ (by orthonormality) and $\|v-p\|^2\ge0$. Hence
$$\|v\|^2\ge\|p\|^2=\sum_i|\langle v,e_i\rangle|^2.$$

Conclusion. Bessel's inequality holds. Equality iff $v\in\operatorname{span}(e_1,\ldots,e_k)$.