= Solution
Let $L$ be a <splitting field> of $f(X)=X^m-1$ over $k$. Its <formal derivative> is
$$
f'(X)=mX^{m-1}.
$$
The assumption on the <characteristic> of $k$ says that $m\ne0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f'(\alpha)=m\alpha^{m-1}\ne0$. Thus $f$ and $f'$ have no common root, so $f$ is a <separable polynomial>. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the <separability of roots of unity>.
Fix a <primitive root of unity> $\zeta_m=e^{2\pi i/m}$. Define the $m$th <cyclotomic polynomial> by
$$
\boxed{\Phi_m(X)=
\prod_{\substack{1\leq a\leq m\\(a,m)=1}}
(X-\zeta_m^a).}
$$
It is <monic> and has degree given by the <Euler totient function> $\varphi(m)$. Partitioning all $m$th <roots of unity> according to their exact <multiplicative order> gives the <cyclotomic factorization>
$$
X^m-1=\prod_{d\mid m}\Phi_d(X).
$$
We now prove by <mathematical induction> that $\Phi_m\in\mathbb Z[X]$. The base case is $\Phi_1=X-1$. If the result holds for every proper divisor $d$ of $m$, then
$$
G(X)=\prod_{\substack{d\mid m\\d<m}}\Phi_d(X)
$$
is monic and belongs to $\mathbb Z[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb Z[X]$. The factorization over $\mathbb C$ says that the remainder is zero and the quotient is $\Phi_m$, proving the claim.
It remains to prove <irreducibility of cyclotomic polynomials>. Let $f\in\mathbb Z[X]$ be a monic <irreducible polynomial> dividing $\Phi_m$, let $\varepsilon$ be a root of $f$, and write $\Phi_m=fg$. We claim that $\varepsilon^p$ is a root of $f$ for every <prime number> $p\nmid m$. Otherwise $\varepsilon^p$ is a root of $g$, so $\varepsilon$ is a root of $g(X^p)$. Since $f$ is the <minimal polynomial> of $\varepsilon$ over $\mathbb Q$, <Gauss lemma for polynomials> gives
$$
f(X)\mid g(X^p)\qquad\text{in }\mathbb Z[X].
$$
After <reduction of an integer polynomial modulo a prime>, the <Frobenius endomorphism> gives
$$
\bar f(X)\mid \bar g(X^p)=\bar g(X)^p.
$$
Choose an irreducible factor $h$ of the nonconstant monic polynomial $\bar f$. Then $h\mid\bar g$, so $h^2\mid\bar f\bar g=\bar\Phi_m$. The <cyclotomic factorization> would then make $h^2$ divide $X^m-1$ over $\mathbb F_p$. This is impossible because its derivative $mX^{m-1}$ is coprime to it when $p\nmid m$.
Thus $f(\varepsilon^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f(\varepsilon^a)=0$ whenever $(a,m)=1$. These are all $\varphi(m)$ primitive $m$th roots, so
$$
\deg f\geq\varphi(m)=\deg\Phi_m.
$$
Because $f$ divides $\Phi_m$, equality holds and $f=\Phi_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb Q$.
Finally, direct multiplication gives
$$
(X^2-X+1)F=X^{10}-X^5+1.
$$
Also
$$
X^{15}+1=(X^5+1)(X^{10}-X^5+1).
$$
Applying the <cyclotomic factorization> to the two factors on the right gives
$$
X^{15}+1=\Phi_2\Phi_6\Phi_{10}\Phi_{30},
\qquad
X^5+1=\Phi_2\Phi_{10},
$$
and therefore
$$
X^{10}-X^5+1=\Phi_6(X)\Phi_{30}(X).
$$
Since $X^2-X+1=\Phi_6(X)$, cancellation yields
$$
\boxed{F(X)=\Phi_{30}(X).}
$$
The <thirtieth cyclotomic polynomial> is irreducible over $\mathbb Q$ by the theorem just proved, so $F$ is irreducible.
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