I will define the metric of $S^{n-1}$ via pullback of the Euclidean metric on ${\mathbb{R}}^{n}$.

To start with we take $n$-dimension Cartesian co-ordinates:
$(x_1,x_2......x_n)$.
The metric here is $g_{ij }= \delta_{ij}$, where $δ$ is the Kronecker delta.

We specify the surface patches of $S^{n-1}$ by the parametrization $f$:
$$x_1=r{\cos{\phi_1}},$$

$$x_p=r{\cos{\phi_p}}{\Pi_{m=1}^{p-1}}{\sin{\phi_{m}}},$$

$$x_n=r{\prod_{m=1}^{n-1}}{\sin{\phi_{m}}},$$

Where $r$ is the radius of the hypersphere and the angles have the usual range.

We see that the pullback of the Euclidean metric $g'_{ab} = (f^*g)_{ab}$ is the metric tensor of the hypersphere. Its components are:

$$g'_{ab} = g_{ij} {\frac{\partial{x_i}}{\partial{\phi_a}}} {\frac{\partial{x_j}}{\partial{\phi_b}}} = {\frac{\partial{x_i}}{\partial{\phi_a}}}{\frac{\partial{x_i}}{\partial{\phi_b}}}$$

We get $2$ cases here:

i) $a>b$ or $b>a$, For these components one obtains a series of terms with alternating signs which vanishes, $g'_{ab}=0$ and thus all off-diagonal components of the tensor vanish.

ii) $a=b$,

$$g'_{11}=1$$

$$g'_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$

where $2\leq a\leq {n-1}$

The determinant is very straightforward to calculate:

$$ \det{(g'_{ab})} = {r^2} \prod_{m=1}^{n-1} g'_{mm}$$

Finally, we can write the metric of the hypersphere as:

$$g' = {r^2} \, d\phi_{1}\otimes d\phi_{1} + {r^2} \sum_{a=2}^{n-1} \left( \prod_{m=1}^{a-1} \sin^2{\phi_{m}} \right) d\phi_{a} \otimes d\phi_{a} $$