How can I tell if $x^5 - (x^4 + x^3 + x^2 + x^1 + 1)$ is/is not part of the solvable group of polynomials?

According to Maple's galois function, the polynomial $x^n - (x^{n-1} + \ldots + 1)$ has Galois group $S_n$ for $n$ from $3$ to $9$ (it can't handle polynomials of degree greater than $9$). In particular, for $n = 5$ to $9$ these are not solvable by radicals. I suspect that none of the polynomials for $n \ge 5$ are solvable by radicals.

You can test its Galois group using this online Magma calculator. For example, to test the solvable but irreducible quintic $x^5-5x+12=0$, use the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

Copy and paste. One then finds the order is $10$, hence that quintic is solvable. (All groups with order $<60$ is solvable, but there are solvable groups with order $>60$.)

For the pentanacci,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-x^4-x^3-x^2-x-1; G, R := GaloisGroup(f); G;

It says it is the symmetric group $S_5$. And for higher $n$-nacci (I tried up to $n=12$), we get $S_n$ which are not solvable for $n\geq5$.

P.S. If you are testing other equations, don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x. (I learned that after a while.)

A result of Dedekind says that for any polynomial $p \in \mathbb{Z}[x]$ and any prime $q$ not dividing the discriminant of $p$, then if $p$ factors modulo $q$ into a product of irreducible polynomials with degrees $d_1, \ldots, d_r$, then the Galois group $\text{Gal}(p)$ contains a permutation with cycle structure $(d_1, \ldots, d_r)$.

The discriminant of the Pentanacci polynomial $p_5(x) := x^5 - (x^4 + x^3 + x^2 + x + 1)$ is $9584 = 2^4 \cdot 599$. It is irreducible modulo $5$ and so $\text{Gal}(p_5)$ contains a $5$-cycle. Modulo $3$, we have $$p_5(x) \equiv (x^3 + x^2 + 2 x + 1) (x^2 + x + 2),$$ so $\text{Gal}(p_5)$ contains a product $\sigma$ of a $2$-cycle and a $3$-cycles and thus also the $2$-cycle $\sigma^3$. Now, if $r$ is prime, then a $2$-cycle and an $r$-cycle in $S_r$ together generate all of $S_r$, and in particular the Galois group of $p_5$ is $S_5$. Now, $S_n$ is only solvable iff $n \leq 4$, so the roots of $p_5$ cannot be extracted with radicals.

Tito and Robert's answers used a dedicated CAS command to show that $\text{Gal}(p_n) \cong S_n$ for $n \leq 12$, and we can readily use Dedekind's result to extend this to, say, $n \leq 20$.

For the $13$-nacci polynomial $p_{13}$, applying an argument similar to the $n = 5$ case, and now instead considering the primes $p = 5$ (which gives a $13$-cycle) and $17$ (which gives an element $\sigma$ with $\sigma^{11}$ a transposition), leads to the conclusion that $\text{Gal}(p_{13}) \cong S_{13}$.

For the $14$-nacci polynomial $p_{14}$, factoring modulo $5$ gives a product $\sigma$ of a $4$-cycle and a $9$-cycle, so $\sigma^{18}$ is a transposition, and factoring modulo $19$ gives a $13$-cycle. But $p_{14}$ is irreducible (as it is irreducible modulo $5$), and so its Galois group is transitive, and being a transitive subgroup of $S_n$ that contains a $2$-cycle and an $(n - 1)$-cycle, $\text{Gal}(p_{14})$ is necessarily $S_{14}$ itself.

We can handle the remaining cases similarly. As above, $\sigma$ is an element with cycle structure given by Dedekind's result for the corresponding prime: $$\begin{array}{c|l} n & p \\ \hline 15 & 11 \, (\sigma^5 \text{ a $2$-cycle}), \, 199 \, (\sigma \text{ a $14$-cycle}) \\ 16 & 5 \, (\sigma^6 \text{ a $2$-cycle}), \, 59 \, (\sigma \text{ a $15$-cycle}) \\ 17 & 3 \, (\sigma^{36} \text{ a $2$-cycle}), \, 5 \, (\sigma \text{ a $17$-cycle}) \\ 18 & 5 \, (\sigma^{26} \text{ a $2$-cycle}), \, 17 \, (\sigma \text{ a $17$-cycle}) \\ 19 & 3 \, (\sigma \text{ a $18$-cycle}), \, 13 \, (\sigma^{33} \text{ a $2$-cycle}) \\ 20 & 5 \, (\sigma^{66} \text{ a $2$-cycle}), \, 23 \, (\sigma \text{ a $19$-cycle}) \end{array}$$