Show that the matrix respect to $T^*$ is the same as the conjugate transpose of the matrix of $T$

Let $V$ be a Hilbert space with orthonormal basis $e_1,e_2,\ldots$. Let $T:V\to V$ be a linear transformation. Let $A$ be the matrix of $T$ with respect to the basis $e_1,e_2,\ldots$ and let $B$ be the matrix of $T^*$ with respect to the basis $e_1,e_2,\ldots$.

Show that $B=\overline A^t$.

I know that $\langle T(v),v\rangle=\langle v,T^*(v)\rangle$ and I tried to expand it. It turns out that $$a_1\langle A_1,v\rangle+a_2\langle A_2,v\rangle+\cdots+a_n\langle A_n,v\rangle=\overline a_1\langle v,B_1\rangle+\overline a_2\langle v,B_2\rangle+\cdots+\overline a_n\langle v,B_n\rangle$$ where $A_i = i$-th column of $A$ and $B_i = i$-th column of $B$

I'm not sure if I'm on the right track and can someone show me the steps to solve this question?

First of all, to be able to talk about matrices of linear transformations, your map $T : V \to V$ has to be bounded. This also assures that $T^*$ exists and is bounded, so it also has a matrix.

Let $A = (a_{ij})$ be the matrix of $T$. The coefficients $a_{ij}$ are defined so that for any $j \in \mathbb{N}$ we have:

$$Te_j = \sum_{i=1}^\infty a_{ij}e_i \implies \langle Te_j, e_i\rangle = a_{ij}$$

Similarly, let $B = (b_{ij})$ be the matrix of $T^*$. The coefficients $b_{ij}$ are defined so that for any $j \in \mathbb{N}$ we have:

$$T^*e_j = \sum_{i=1}^\infty b_{ij}e_i \implies \langle T^*e_j, e_i\rangle = b_{ij}$$

Now for any $i, j \in \mathbb{N}$ we have:

$$b_{ij} = \langle T^*e_j, e_i\rangle = \langle e_j, Te_i\rangle = \overline{\langle Te_i, e_j\rangle} = \overline{a_{ji}}$$

Therefore $B = A^* = \overline{A^t}$, i.e. $B$ is the conjugate transpose of $A$.