Prove that $\det(\varphi(\sigma)) = \operatorname{sgn}(\sigma)$ for all $\sigma \in S_n$. 
Fix a positive integer $n$.
For each $\sigma \in S_n$ define an $n\times n$ matrix, $\varphi(σ)$ by
$$\varphi (\sigma)_{i,j}= \begin{cases} 1 & \text{if } i = \sigma (j), \\ 0 & \text{if } i \neq \sigma (j). \end{cases}
$$

I have proved that $\varphi$ is a homomorphism and found $\ker \varphi.$

Prove that $\det(\varphi(σ)) = \operatorname{sgn}(σ)$, for all $σ \in S_n$

There should be some nice way to do this considering that $\det,$ $\operatorname{sgn},$ and $\varphi$ are all homomorphisms. 
I tried writing it out for a single transposition but wasn't sure how to generalize it?
I want to link it to this question Prove that for each $\phi \in S_n$, $\det A=\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\prod_{j=1}^n a_{\phi(j)\sigma(j)}$ cause it feels very similar but i cant seem to figure out how (and i dont understand the answer given here)
 A: If $\sigma$ is a product of $k$ transpositions, then $\operatorname{sgn}(\sigma) = (-1)^k,$ so if $\sigma$ is a transposition, then $\operatorname{sgn}(\sigma) = -1.$
Suppose $\sigma = \tau_1 \circ \cdots \circ \tau_k$ where $\tau_1,\ldots,\tau_k$ are transpositions.
Then
\begin{align}
& \overbrace{ \varphi(\tau_1)\cdots \cdots\varphi(\tau_k) = \varphi(\sigma) }^{\Large\text{since $\varphi$ is a homomorphism}}
\end{align}
and therefore
\begin{align}
\det \varphi\left(\sigma\right)
= (\det \varphi(\tau_1)) \cdots (\det \varphi(\tau_k))
= (-1)^k
\end{align}
provided that one can show $\det(\varphi(\tau_i)) = -1$ whenever $\tau_i$ is a transposition. But all we need for that is that the determinant of the identity matrix is $1$ and $\varphi(\tau_i)$ comes from interchanging two rows of the identity matrix, and interchanging two rows multiplies the determinant by $-1.$
Then we have
$$
\varphi(\rho\sigma) = (-1)^{\text{a certain power}}
$$
and you can think about what that exponent is.
A: $\operatorname{sgn}$ is by definition the only nontrivial morphism $\mathfrak{S}_n \to \mathbb{C}^{\times}$.
${\det}\circ \phi$ is a morphism $\mathfrak{S}_n \to \mathbb{C}^{\times}$. Clearly it is nontrivial, so it is $\operatorname{sgn}$.
