We know that given a square matrix $A$, we can find its eigenvalues through the characteristic polynomial:
$$\text{det}(A - \lambda I) = \lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0$$
Suppose I wish to factorize this polynomial in term of the $n$ roots of this polynomial, which I denote as $\lambda_1, \ldots, \lambda_n$, which of the following answer is correct?
(a) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (\lambda - \lambda_i)$
(b) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (-1)^n(\lambda - \lambda_i)$
(c) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (\lambda_i - \lambda)$
(d) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (-1)^n (\lambda_i - \lambda)$
I am a bit confused because $(-1)^n$ may produce a leading coefficient of $-1$ whenever $n$ is odd.
To expand a little bit on my comment:
The correct expansion of $\det(A-\lambda I)$ is $(-1)\lambda^n+c_{n-1}\lambda^{n-1}+\dots$. However, in the subsequent question, you are trying to factor the monic polynomial $\lambda^n+c_{n-1}\lambda^{n-1}+\dots$.
In general, a polynomial $P=c_n\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0$ has factorizarion $c_{n}\prod_{i=1}^n (\lambda-\lambda_i)$, where $\lambda_i$ are the roots of $P$.
The answer a) is thus correct, since here $c_n=1$.
But there is a trap.
b) is wrong in general because, if you get the $(-1)^n$ factor out of the product, you have then $(-1)^{n^2}\prod_{i=1}^n (\lambda-\lambda_i)$, and $(-1)^{n^2}\ne1$ for odd $n$.
The answer c) is also wrong, because by changing the sign of each factor and getting a factor $(-1)$ out, you have really $(-1)^{n}\prod_{i=1}^n (\lambda-\lambda_i)$, and again $(-1)^n\ne1$ if $n$ is odd.
And here comes the trap. For answer d), you can rewrite the product as $(-1)^{n^2+n}\prod_{i=1}^n (\lambda-\lambda_i)$, and $n^2+n=n(n+1)$ is always even, therefore it's really $\prod_{i=1}^n (\lambda-\lambda_i)$, and answer d) is correct.
However, if your goal is to factor $\det(A-\lambda I)$, and the eigenvalues are $\lambda_i$, i.e. these are the roots of the polynomial $\det(A-\lambda I)$, then you can write:
$$\det(A-\lambda I)=(-1)^n \prod_{i=1}^n (\lambda-\lambda_i)$$
Here the polynomial $\det(A-\lambda I)$ is not monic, it has leading term $(-1)^n\lambda^n$, as can be seen by expanding the determinant by Leibniz formula: the only term in the expansion with $n$ factors $\lambda$ comes from the identity permutation (and its signature is $+1$), i.e. it's $\prod_{i=1}^n (a_{ii}-\lambda)$, and when you expand this product the only term of degree $n$ is $(-1)^n\lambda^n$.