The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ \langle a V_1 + b V_2, c U_1 + d U_2 \rangle = ac \langle V_1, U_2\rangle + ad \langle V_2, U_1 \rangle + bc \langle V_2, U_1 \rangle + bd \langle V_2, U_2 \rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$\langle e_j, e_k \rangle = \Big\langle \frac{v_j - \langle v_j, e_1 \rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1}}{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}, e_k\Big\rangle$$
$$= \frac{\Big\langle v_j - \langle v_j, e_1 \rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1}, e_k\Big\rangle}{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
$$= \frac{\Big\langle v_j, e_k\Big\rangle - \langle v_j, e_1 \rangle \Big\langle e_1, e_k\Big\rangle - \dots - \langle v_j, e_{k}\rangle \Big\langle e_{k}, e_k\Big\rangle \dots- \langle v_j, e_{j-1}\rangle \Big\langle e_{j-1}, e_k\Big\rangle}{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
By construction the basis vectors $e_1 \dots e_k \dots e_{j-1}$ are ortho-normal. This means that $\langle e_i, e_k \rangle = 0$ when $i \neq k$ and $\langle e_k, e_k \rangle = 1$ for $1\leq i \leq j-1$.
$$= \frac{\Big\langle v_j, e_k\Big\rangle - \langle v_j, e_1 \rangle 0 - \dots - \langle v_j, e_{k}\rangle 1 \dots- \langle v_j, e_{j-1}\rangle 0}{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
$$= \frac{\Big\langle v_j, e_k\Big\rangle - 0 - \dots - \langle v_j, e_{k}\rangle \dots- 0}{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
$$= \frac{\Big\langle v_j, e_k\Big\rangle - \langle v_j, e_{k}\rangle }{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
$$= \frac{0 }{\| v_j - \langle v_j, e_1\rangle e_1 - \dots - \langle v_j, e_{j-1}\rangle e_{j-1} \|}$$
$$= 0$$
Best Answer
Use the fact that the zero vector is the neutral element for vector addition, so it is equal to itself minus itself. Then use the distributive law for scalar multiplication.