Proving the irreducibility of a specific family of polynomials I want to show that $f(x)=x^{4k} - 3x ^{3k} + 4x^{2k}-2x^k +1$ is irreducible in $\mathbb{Q}$ for all $k\in \mathbb{N}$. When $k=1$, it is easy to show; however I have trouble in proving this while $k\ge 2$. I have tried lots of irreducibility tests, but I have not found a way to prove this. Can anyone give me, at least, a hint?

Lemma: If $F$ contains a primitive $k$th root of unity then $f(x)=x^k-b$ is irreducible over $F$ if $b$ has not any $n$th root in $F$, $n>1$.
Proof: We know $A=\{\sqrt[k]{b},w\sqrt[k]{b},w^2\sqrt[k]{b},...,w^{k-1}\sqrt[k]{b}\}$ is a subset of $K=F(\sqrt[k]{b})$ so $K/F$ is Galois. Its Galois group is a subgroup of $\mathbb Z_k$ because the roots of minimal polynomial of $\sqrt[k]{b}$ are in $A$, so $\phi:G\to \mathbb Z_k:\phi(\eta)=i$ if $\eta(\sqrt[k]{b})/\sqrt[k]{b}=w^i$ is an injective homomorphism. If $g$ is the minimal polynomial of $\sqrt[k]{b}$, then $g(0)=\prod_{j\in G}{w^j\sqrt[k]{b}}=\sqrt[k]{b^{\deg(g)}}$, so $g(0)\in F \iff \deg(g)=k$, so $g=f$ and $f$ is irreducible.
Let $f=x^{4k}-3x^{3k}+4x^{2k}-2x^k+1$. To prove $f$ is irreducible it is sufficient to show $[K:\mathbb Q]=4k$ where $K=\mathbb Q(\sqrt[k]{1+e^{2\pi i/5}})$. By using the tower lemma we have $$[K:\mathbb Q]=[K:F][F:\mathbb Q]=4[K:F]\ (F=\mathbb Q(e^{2\pi i/5}))$$ so it is sufficient to show $[K:F]=k$ or $x^k-(1+e^{2\pi i/5})$ is irreducible over $F$. But $1+e^{2\pi i/5}$ hasn't any $n$th root in $F$, so by the lemma it is irreducible over $F(w)$, so over $F$. 
