Proving the determinant of a square matrix $A$ is the same as the determinant of it's jordan matrix, $J$

Question

I am wondering if this follows directly from the fact that an $(n\times n)$ matrix $A$ can be written as $$ \tag{1} A = MJM^{-1} $$ with $M$ being $(n\times n)$ and nonsingular, and $J$ being the Jordan Matrix of $A$ (the matrix with the eigenvalues of $A$ on principal diagonal and ones or zero on diagonal above principle diagonal), or if the result in the title requires using the fact that an eigenvalue of $A$ is also an eigenvalue of $SAS^{-1}$ for any nonsingular $S$.

To me it seems like it follows from $(1)$ because $$ \lvert A\rvert = \lvert MJM^{-1}\rvert = \lvert MJ\rvert\lvert M^{-1}\rvert = \lvert M \rvert \lvert J\rvert \frac{1}{\lvert M \rvert} = \lvert J\rvert $$ but I wish to confirm.

Also, if I do need to use the fact that an eigenvalue of $A$ is also an eigenvalue of $SAS^{-1}$ for any nonsingular $S$, could I get a hint or an explanation of where/why this is required?

Thanks.

Answer 1

Question was answered in the comments so I'll put it as an answer:

The result follows directly from $(1)$: I do not need to use the fact that an eigenvalue of $A$ is also an eigenvalue of $SAS^{-1}$ for any nonsingular $S$.