Galois group of $x^5+2$ over $\mathbb{Q}$

I want to find the Galois group $G$ of $f=x^5+2$ over $\mathbb{Q}$. I will write my approach, and I would like to know if there is a faster approach or a more standard one that does not require creativity.

Let $\alpha$ be such that $\alpha^5=-2$ and $\zeta$ be the fifth root of unity. Then, the roots of $f$ are $\{\alpha,\zeta\alpha,\zeta^2\alpha,\zeta^3\alpha,\zeta^4\alpha\}$. The splitting field of $f$ is $\mathbb{Q}(\alpha,\zeta)$. First, I want to find the degree $$[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=5[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}(\alpha)].$$ This gives me 3 possibilities for the degree: $5$, $10$, and $20$. Now, to find this degree and the Galois group $G$ simultaneously, I use the fact that the Galois group of $f$ over finite fields embeds into the Galois group $G$ over $\mathbb{Q}$.

Doing everything mod $3$, we get $$f=x^5-1=(x-1)(1+x+x^2+x^3+x^4)$$ with roots $\{1,\zeta,\zeta^2,\zeta^3,\zeta^4\}$. If $\sigma:x\mapsto x^3$ be the Frobenius map, then $$\sigma(\zeta)=\zeta^3 \qquad \sigma^2(\zeta)=\zeta^9=\zeta^4 \qquad \sigma^3(\zeta)=\zeta^{12}=\zeta^2 \qquad \sigma^4(\zeta)=\zeta.$$ This tells me $1+x+x^2+x^3+x^4$ is irreducible over $\mathbb{F}_3$, and its Galois group is $\mathbb{Z}/4\mathbb{Z}$, which is the same for $f$. Consequently, $\mathbb{Z}/4\mathbb{Z}$ embeds into the $G$, and so the only possibility for the degree $|G|$ is $20$. Moreover, $G$ must have element of order $5$, so $G=\mathbb{Z}/5\mathbb{Z}\ltimes \mathbb{Z}/4\mathbb{Z}$. Probably, $G=\langle (12345),(2345)\rangle$, but I am not sure.

Is everything right? Do you suggest a better or faster approach?