Proof Verification: Abelian groups have elements with orders equal to each of the group's prime divisors

Question

I was self-studying out of Aluffi's Algebra: Chapter 0, and I came across this exercise:

Aluffi, Algrbra: Chapter 0, Exercise II.8.17

Assume $G$ is a finite abelian group, and let $p$ be a prime divisor of $|G|$. Prove that there exists an element in $G$ of order $p$. (Hint: let $g$ be any element of $G$, and consider the subgroup $\langle g\rangle$; use the fact that this subgroup is cyclic to show that there is an element $h\in\langle g\rangle$ in $G$ of prime order $q$. If $q = p$ you are done; else, use the quotient $G/\langle h\rangle$ and induction.)

My question:

I feel fine with most of my proof (although it may be a bit drawn out), but I haven't quite convinced myself with the final paragraph, after the induction. I could say in the induction step, "If q=p, we're done. If not, move on to the next step," but this reminds of of the sort of 'algorithmic proofs' seen in the earlier sections of Axler's Linear Algebra Done Right, which I feel are of questionable rigor. Is there a better way to express the idea I'm going for in the final paragraph, is there a better idea altogether, or is my treatment just fine?

My proof:

Let $G_0=G$. We will soon define each $G_k$ to be a quotient $G_{k-1}/H$, where $H$ is a subgroup of prime order. Let $\Omega(n)$ be the number of prime divisors of $n$, including multiplicity. We will proceed by (strong) induction on $k$, proving that, for $0\leq n < \Omega(|G|)$:

1. $\Omega(|G_n|) = \Omega(|G|) - n$
2. There exists an element of $|G_n|$ of prime order.
3. If $g\in G_n$ has prime order $p$, then there exists an element of $G$ of order p.

Base Case (n=0): Let $G_0=G$.

1. $\Omega(|G_0|) = \Omega(|G|) - 0$
2. Let $g_0\in G_0$, and for some prime divisor $q_0$ of |g_0|, let $h_0=g_0^{\frac{|g_0|}{q_0}}$. We then have $|h_0| = q_0$, so (2) holds.
3. Trivial.

Induction (n=k+1): Suppose (1), (2), and (3) hold for $n\leq k$. Let $H_k = \langle h_k\rangle$ (which is normal because $G$ is abelian), and define $G_{k+1}=G_k/H_k$.

1. We have: $$ \begin{align*} \Omega(|G_{k+1}|) &= \Omega\left(\left|\frac{G_k}{H}\right|\right) \\ &= \Omega\left( \frac{|G_k|}{|H|} \right) \\ &= \Omega\left( \frac{|G_k|}{q} \right) \\ &= \Omega(|G_k|) - 1 \\ &= \Omega(G) - k - 1 \end{align*} $$
2. Same as the base case - let $q_{k+1}$ be a prime divisor of $|G_{k+1}|$, let $g_{k+1}\in G_{k+1}$, and define $h_{k+1}=g^{\frac{|g_{k+1}|}{q_{k+1}}}$. Then $|h_{k+1}|=q_{k+1}$.
3. Suppose $gH_k\in G_{k+1}$ has order $p$, where $p$ is a prime divisor of $|G_{k+1}|$. It then follows that $(gH_k)^p = e$, and so $g^p=h_k^m$ for some integer $m$. Since $e=(h^k)^x = g^{px}$ for some $x=|h^k|$, we then have $|g^x|=p$ in $G_k$. By our inductive hypothesis, then, we have an element of $G$ of order $p$, as desired.

The set $\{q_0, \dots, q_n\}$ where $n=\Omega(G)-1$ is the set of all prime divisors of $|G|$, since at each step of the process we have removed a prime divisor before choosing a new one. At the end of this process, we have proven that $G$ contains an element of order $q_n$ for each $n$, and hence an element of order $p$ for each prime divisor $p$ of $|G|$.