Degrees of the irreducible factors of $X^{13}-1$ I need some hints to determine the degrees of irreducible factors of $X^{13}-1$ in $\Bbb F_5[X]$. I know the first factor is $X-1$ and it is a single root. There are not other roots. I also checked from Mathematica that $1+X+ \dots X^{12}$ is not irreducible in $\Bbb F_5[X]$ but I lack proper argument to show it.
I know by Mathematica that $X^{13}-1$ can be factored as 
$$(X+4) \left(X^4+X^3+4 X^2+X+1\right) \left(X^4+2 X^3+X^2+2 X+1\right) \left(X^4+3 X^3+3 X+1\right)$$
How do I determine the degrees without Mathematica?

We know that $X^{5^n} - X$ is the product of all irreducible polynomials over $\mathbb{F}_5$ whose degree divides $n$. You can compute
$$\mathrm{gcd}(X^{13} - 1, X^5 - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{25} - X) = X - 1,$$
$$\mathrm{gcd}(X^{13} - 1, X^{625} - X) = X^{13} - 1$$ with the Euclidean algorithm and immediately read off that $X^{13} - 1$ has one factors of degree $1$, no factors of degree $2$ and $\frac{12}{4} = 3$ factors of degree $4$.
The fact that $X^{13} - 1$ divides $X^{625} - X$ mod $5$ is because $13 \mid 624$, as suggested in the comments.
