Characteristic value or eigenvalues and determinant I am having semester in linear algebra. And have recently got acquainted to eigenvalues.

What is the relation between eigenvalues and determinant? Going through answers of some questions I found suggestion of using eigenvalues to determine determinant of matrix. I just know $\det(eI-A)=0$ where e is eigenvalue of A. How to use this fact to calculate determinant of A?
You know that the eigenvalues are the roots (zeros) of the characteristic polynomial, i.e. of the polynomial $(x) = \det(xI - A)$. Watch what happens of you evaluate that polynomial at $0$, i.e. compute $p(0)$. You get $p(0) = \det(0I-A) = \det(-A)$. Now, $\det(-A) = (-1)^n\det A$ (assuming that A is a $n\times n$ matrix).

So how do you find $p(0)$, if you only have the eigenvalues of $A$. Well, since the eigenvalues of $A$ are the zeros of $p(x)$, you can write $p(x)$ as a product $$
  p(x) = \alpha(x - \lambda_1)(x - \lambda_2)\ldots(x - \lambda_n)
$$
where $\lambda_1,\ldots,\lambda_n$ are the zeros of $p(x)$, i.e. the eigenvalues of $A$. Note that some particular $\lambda$ will appear more than once in this list, if $p(x)$ has a zero with an order higher than 1 at $\lambda$. For characteristic polynomials, you also know that $\alpha = 1$, because the coefficient of $x^n$ in the characteristic polynomial of a $n\times n$ matrix is always $1$. Now watch what happens if you evaluate $p$ at $0$. You get $$
  p(0) = (0 - \lambda_1)(0 - \lambda_2)\ldots(0 - \lambda_n) = (-1)^n\lambda_1\ldots\lambda_n \text{.}
$$

Combining these results, you have $$
  p(0) = (-1)^n\det A = (-1)^n\lambda_1\ldots\lambda_n
  \implies
  \det A = \lambda_1\ldots\lambda_n \text{.}
$$
Write the characteristic polynomial $\chi_A$ as $\chi_A(X)=\prod_{\lambda\text{ eigenvalue}}(X-\lambda)$; what is $\chi_A(0)$? You can express it as the constant coefficient (easy to compute, function of the eigenvalues only), or with your expression above (taking your "$e$" to be 0))

(in the product, the $\lambda$'s are not necessarily distinct)