I am a beginning Linear Algebra learner and this is just my humble opinion.

One idea presented above is that

Suppose that $\lambda_1,\ldots \lambda_2$ are eigenvalues of $A$.

Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.

$$\det(A−\lambda I)=(\lambda_1-\lambda)(\lambda_2−\lambda)\cdots(\lambda_n−\lambda)$$.

Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1\lambda_2\ldots \lambda_n$, that is, we indeed obtain the desired result

$$\det(A)=\lambda_1\lambda_2\ldots \lambda_n$$.

I dont think that this works generally but only for the case when $\det(A) = 0$.

Because, when we write down the characteristic equation, we use the relation $\det(A - \lambda I) = 0$ Following the same logic, the only case where $\det(A - \lambda I) = \det(A) = 0$ is that $\lambda = 0$.
The relationship $\det(A - \lambda I) = 0$ must be obeyed even for the special case $\lambda = 0$, which implies, $\det(A) = 0$

**UPDATED POST**

Here i propose a way to prove the theorem for a 2 by 2 case.
Let $A$ be a 2 by 2 matrix.

$$ A = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\\end{pmatrix}$$

The idea is to use a certain property of determinants,

$$ \begin{vmatrix} a_{11} + b_{11} & a_{12} \\ a_{21} + b_{21} & a_{22}\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\\end{vmatrix} + \begin{vmatrix} b_{11} & a_{12}\\b_{21} & a_{22}\\\end{vmatrix}$$

Let $ \lambda_1$ and $\lambda_2$ be the 2 eigenvalues of the matrix $A$. (The eigenvalues can be distinct, or repeated, real or complex it doesn't matter.)

The two eigenvalues $\lambda_1$ and $\lambda_2$ must satisfy the following condition :

$$\det (A -I\lambda) = 0 $$
Where $\lambda$ is the eigenvalue of $A$.

Therefore,
$$\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = 0 $$

Therefore, using the property of determinants provided above, I will try to *decompose* the determinant into parts.

$$\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} - \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix}= \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\\\end{vmatrix} - \begin{vmatrix} a_{11} & a_{12} \\ 0 & \lambda \\\end{vmatrix}-\begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix}$$

The final determinant can be further reduced.

$$
\begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} \\\end{vmatrix} - \begin{vmatrix} \lambda & 0\\ 0 & \lambda\\\end{vmatrix}
$$

Substituting the final determinant, we will have

$$
\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\\\end{vmatrix} - \begin{vmatrix} a_{11} & a_{12} \\ 0 & \lambda \\\end{vmatrix} - \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} \\\end{vmatrix} + \begin{vmatrix} \lambda & 0\\ 0 & \lambda\\\end{vmatrix} = 0
$$

In a polynomial
$$ a_{n}\lambda^n + a_{n-1}\lambda^{n-1} ........a_{1}\lambda + a_{0}\lambda^0 = 0$$
We have the product of root being the coefficient of the term with the 0th power, $a_{0}$.

From the decomposed determinant, the only term which doesn't involve $\lambda$ would be the first term

$$
\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\\end{vmatrix} = \det (A)
$$

Therefore, the product of roots aka product of eigenvalues of $A$ is equivalent to the determinant of $A$.

I am having difficulties to generalize this idea of proof to the $n$ by $$ case though, as it is complex and time consuming for me.