Take $G$ to be a group of order $600$. Prove that for any element a in $G$ there exist a $b$ in $G$ such that $a = b^7$

Take $G$ to be a group of order $600$. Prove that for any element $a$ $\in$ G there exist an element $b$ $\in$ G such that $a = b^7$.

My thought process: since $a = b^7$ $\implies$ $|a| = |b^7|$. Consequently $\operatorname{lcm}(1,|a|) = \dfrac{1}{7}\operatorname{lcm}(7,|b|)$ implies $7|a| = 7|b|$ so $|a| = |b|$. I don't know where I would go from here or if this is even the right approach.

Take the cyclic subgroup generated by $a$. It has some order $k$ so $a^{1+kn}=a$ for all $n$. Let $b=a^l$

$$ b^7=a^{1+kn}\\ a^{7l}=a^{1+kn}\\ $$

$7l \equiv 1 \; (mod \; k)$

Suppose $k=2$, then $l=1$ works and $b=a$ satisfies $b^7=b=a$.

Suppose $k=20$, then $l=3$ work $b=a^3$ satisfies $b^7=a^{21}=a$

What are the possible $k$ and what can you then say about the solvability of $7l \equiv 1$?

Hint: $\gcd(600,7)=1 \implies 1=600m+7n$. Indeed, $ 1 = 3 \cdot 600 - 257 \cdot 7$.

Proposition Let $G$ be a finite group and $n$ a positive integer. Then the map $f: G \mapsto G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(|G|,n)=1$.

Proof (sketch) Bézout yields $1=k|G|+mn$, for some integers $k, m$. Then $g=g^{k|G|+mn}=g^{mn}$. Hence if $g^n=h^n$, then $g=g^{mn}=h^{mn}=h$. So $f$ is injective and since $G$ is finite it must be bijective. Conversely, assume gcd$(n,|G|)\neq 1$. Then we can find a prime $p$ with $p \mid n$ and $p \mid |G|$. By Cauchy's Theorem there is a non-trivial $g \in G$ with order$(g)=p$. Then $g^n=g^{p \cdot \frac{n}{p}}=1^\frac{n}{p}=1=1^n$. Since $f$ is injective this yields $g=1$, a contradiction.