Here is my approach:
If $a\in G$, let's consider a subgroup $H=[a]$. Lagrange's theorem tell us that $o(H) \mid p^n$. But since $p$ is prime, the order $H$ must be the power $p\leq n$, with $k\leq n$. In particular, $o(a)=p^k$
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1We don't need to write $o(H)$ of "de subgroup" anymore, see here. By Lagrange, the order $o(g)$ divides $p^n$, and hence is itself a prime power $p^k$. – Dietrich Burde Feb 16 '22 at 13:36
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Even if $G$ is not cyclic? – Marcos Paulo Feb 16 '22 at 13:47
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A group of prime power order is not cyclic in general. So yes, for all finite groups, see the duplicate in my comment. – Dietrich Burde Feb 16 '22 at 14:00
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Actually when I said "de", I should have written "a". – Marcos Paulo Feb 16 '22 at 19:37
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Yes, I know. Probably it came from a Spanish text? Ah, Portuguese! – Dietrich Burde Feb 16 '22 at 19:43
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Portuguese... I am Brazilian – Marcos Paulo Feb 16 '22 at 19:45
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@DietrichBurde $Z_9$ is group whose order is 9 which is a power of 3 and is cyclic same is true for any $Z_{p^n}$ group. – grey Feb 16 '22 at 21:37
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@grey $C_p\times C_p$ is a group of order $p^2$, but is not cyclic. A group of order $p^3$ even need not be abelian. Think of the Heisenberg group over $\Bbb F_p$. So again: " A group of prime power order is not cyclic in general." – Dietrich Burde Feb 16 '22 at 21:42
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@DietrichBurde what do you mean when you say "in general"? I'm a bit curious about this. – grey Feb 16 '22 at 22:09
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@grey "Not cyclic in general" means, not necessarily cyclic. This is a standard formulation in mathematics. Even integers are not divisible by four in general - means, that an integer like $2$ or $6$ is not divisible by $4$, but $8$ is divisible by $4$, for example. – Dietrich Burde Feb 16 '22 at 22:38