Letting $x$ and $y$ be elements of some separable Hilbert space $\mathcal{H}$, we know that for any orthonormal basis $\{e_i\}_{i=1}^\infty$, we can write
$$ x = \sum_{i=1}^\infty \langle x, e_i \rangle e_i $$
and similarly for $y$. Does it then follow that
$$ \langle x, y \rangle = \left\langle \sum_{i=1}^\infty \langle x, e_i \rangle e_i, \sum_{j=1}^\infty\langle y, e_j \rangle e_j \right\rangle = \sum_{i=1}^\infty \sum_{j=1}^\infty \langle x, e_i \rangle \langle y, e_j \rangle \langle e_i, e_j \rangle = \sum_{i=1}^\infty \langle x, e_i \rangle \langle y, e_i \rangle $$
I'm worried that bilinearity does not hold for infinite sums and I can't find a reference to this result anywhere, so I can't really tell if it is so trivial from Parseval, that noone bothers to mention it or if it is wrong.
The calculation is justified because the inner product is continuous. You can also get the result by noting that, by Bessel's equality, the map $\hat x:\mathcal H \to \ell^2(\mathbb N,\mathbb C):x\mapsto (\langle x,e_n\rangle)^{\infty}_{n=1} $ is an isometry, so an application of the polarization identity gives $\langle x,y\rangle=\langle \hat x,\hat y\rangle$ and the right hand side of this is $\sum_{i=1}^\infty \langle x, e_i \rangle \langle y, e_i \rangle$ by definition of the inner product on $\ell^2(\mathbb N,\mathbb C).$