Orthonormal basis and inner product 
Let $\langle x, y\rangle$ be an inner product on a vector space $V$, and let $e_1, e_2, \cdots, e_n$ be an orthonormal basis for $V$.

Question. 

Prove: $\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = a_1b_1 + a_2b_2 + \cdots + a_nb_n$, where $a_i,b_i\in F$.

Answer attempt.  
Since $\{e_1, e_2, \cdots, e_n\}$ is an orthonormal set, $\langle e_i, e_j\rangle = 0$ when $i \neq j$ and $\langle e_i, e_j\rangle = 1$ when $i = j$. So:
$\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = \langle a_1e_1, b_1e_1\rangle + \langle a_2e_2, b_2e_2\rangle + \cdots + \langle a_ne_n, b_ne_n\rangle = a_1\overline{b_1} + a_2\overline{b_2} + \cdots + a_n\overline{b_n}$
Associated question

Prove: $\langle x, y\rangle = \langle x, e_1\rangle\langle y, e_1\rangle + \langle x, e_2\rangle\langle y, e_2\rangle + \cdots + \langle x, e_n\rangle\langle y, e_n\rangle$

Answer attempt
Similarly, I get: 
$\langle x, y\rangle = \langle x, e_1\rangle\overline{\langle y, e_1\rangle} + \langle x, e_2\rangle\overline{\langle y, e_2\rangle} + \cdots + \langle x, e_n\rangle\overline{\langle y, e_n\rangle}$
...am I missing something? If yes, I'd really appreciate it if someone would point out what.  
 A: The first answer is right, also, note that
$$\begin{align}
\left\langle \sum_{i=1}^n a_ie_i, \sum_{j=1}^n b_je_j \right\rangle
&= \sum_{i=1}^n a_i \left\langle e_i, \sum_{j=1}^n b_je_j \right\rangle \\
&= \sum_{i=1}^n \sum_{j=1}^n a_i\overline{b_j} \langle e_i, e_j \rangle \\
&= \sum_{i=1}^n \sum_{j=1}^n a_i\overline{b_j} \delta_{ij} \\
&= \sum_{i=1}^n a_i\overline{b_i}
\end{align}$$
where $\delta_{ij}$ is the Kronecker's delta: $\delta_{ij} = 1$ if $i=j$ and $\delta_{ij} = 0$ if $i\neq j$.
For the second part, you need some more elaboration. Since $e_1,\dots,e_n$ is orthonormal, every vector $x$ can be written as
$$x = \sum_{i=1}^n \langle x,e_i \rangle e_i$$
and similarly,
$$y = \sum_{j=1}^n \langle y,e_j \rangle e_j.$$
So, applying the previous result with $a_i = \langle x,e_i \rangle$ and $b_j = \langle y,e_j \rangle$ we get the desired result.
A: The second follows immediately from the first, using $x=a_1e_1+\dots+a_ne_n$ and $y=b_1e_1+\dots+b_ne_n$.
There shouldn't be any complex conjugate symbols.  You said inner product, not Hermitian inner product.
