I know splitting fields of finite fields $\mathbb{F}_p$ are of the form $\mathbb{F}_{p^n}$ where n is the degree of the extension over $\mathbb{F}_p$. Consequently, intermediate fields are of the form $\mathbb{F}_{p^d}$ where $d$ divides $n$.
Also, I know the the polynomial $x^{p^n}-x$ is precisely the product of the all the irreducible polynomials of degree that divide $n$, and that its splitting field is $\mathbb{F}_{p^n}$.
Here are my thoughts. $(x^3+x^2+1)(x^3+x+1)(x^2+x+1)$ divides $x^{2^6}-x$. All that is missing are the 9 irreducible polynomials of degree 6, $x+1$ and $x$. However, the last two won't contribute anything to the splitting field. So the splitting field must be contained in $\mathbb{F}_{2^6}$, and so is one of $\mathbb{F}_{2^d}$ where $d = 1, 2, 3, \text{or } 6.$ However it is at least $\mathbb{F}_{2^3}$ as it contains the splitting field of $x^3+x^2+1$ (which is $\mathbb{F}_{2^3}$).
So I just need to rule out the $d=6$ possibility. One way would be present a root of one of the 9 irreducible degree 6 polynomials and show it isn't in the splitting field in question. But that doesn't seem fun. Heck maybe the splitting field is $\mathbb{F}_{2^6}$, I'm not actually sure.
Any tips would be appreciated. Alternate methods appreciated too. Thank you.
(Promoting my comment to answer. I'm fairly sure we have covered essentially the same question, but I don't have time to look for one. Hence CW.)
The splitting field is $\Bbb{F}_{64}$. This is seen as follows:
Those cubic factors need $\Bbb{F}_8$, it must be a subfield of the splitting field.
The quadratic factor needs $\Bbb{F}_4$, so that must also be a subfiled of the splitting field.
$\Bbb{F}_{64}$ is the smallest field that contains both $\Bbb{F}_8$ and $\Bbb{F}_4$ as subfields.