[proofplan] We expand $(v, w)$ by writing both vectors in the orthonormal basis and distributing the inner product using sesquilinearity. The orthonormality condition $(e_i, e_j) = \delta_{ij}$ collapses the double sum to a single sum of products of Fourier coefficients. The special case $w = v$ gives $\|v\|^2 = \sum_i |(e_i, v)|^2$. [/proofplan] [step:Expand $v$ and $w$ in the orthonormal basis] Since $(e_1, \dots, e_n)$ is an orthonormal basis, every vector has a unique expansion. Write $v = \sum_{i=1}^n v_i\,e_i$ and $w = \sum_{j=1}^n w_j\,e_j$. The Fourier coefficients are $v_i = (e_i, v)$ and $w_j = (e_j, w)$, since for each $i$: \begin{align*} (e_i, v) = \Bigl(e_i, \sum_{k=1}^n v_k\,e_k\Bigr) = \sum_{k=1}^n v_k\,(e_i, e_k) = \sum_{k=1}^n v_k\,\delta_{ik} = v_i. \end{align*} [/step] [step:Compute $(v, w)$ and collapse using orthonormality] Using sesquilinearity of the inner product: \begin{align*} (v, w) = \Bigl(\sum_{i=1}^n v_i\,e_i,\; \sum_{j=1}^n w_j\,e_j\Bigr) = \sum_{i=1}^n \sum_{j=1}^n \bar{v}_i\, w_j\, (e_i, e_j). \end{align*} The orthonormality condition $(e_i, e_j) = \delta_{ij}$ annihilates all off-diagonal terms: \begin{align*} (v, w) = \sum_{i=1}^n \bar{v}_i\, w_i = \sum_{i=1}^n \overline{(e_i, v)}\,(e_i, w). \end{align*} Setting $w = v$ gives: \begin{align*} \|v\|^2 = (v, v) = \sum_{i=1}^n \bar{v}_i\, v_i = \sum_{i=1}^n |v_i|^2 = \sum_{i=1}^n |(e_i, v)|^2. \end{align*} [guided] The computation reduces the inner product to coordinates. Starting from $v = \sum_i v_i e_i$ and $w = \sum_j w_j e_j$, the inner product is conjugate-linear in the first argument and linear in the second: \begin{align*} (v, w) = \sum_{i,j} \bar{v}_i\, w_j\, (e_i, e_j). \end{align*} The double sum has $n^2$ terms. However, orthonormality $(e_i, e_j) = \delta_{ij}$ kills all terms where $i \neq j$, leaving only the $n$ diagonal terms: \begin{align*} (v, w) = \sum_{i=1}^n \bar{v}_i\, w_i. \end{align*} Since $v_i = (e_i, v)$ and $w_i = (e_i, w)$ (the Fourier coefficients), we can rewrite this as: \begin{align*} (v, w) = \sum_{i=1}^n \overline{(e_i, v)}\,(e_i, w). \end{align*} This is Parseval's identity: the inner product in $V$ equals the coordinate-wise inner product in $\mathbb{F}^n$. The special case $w = v$ gives $\|v\|^2 = \sum_{i=1}^n |(e_i, v)|^2$, expressing the norm as the sum of squared magnitudes of the Fourier coefficients. This is the finite-dimensional version of the Plancherel theorem: the map $v \mapsto ((e_1, v), \dots, (e_n, v))$ is a unitary isomorphism from $V$ to $\mathbb{F}^n$. [/guided] [/step]