Proof: By Euclid
(related to Proposition: 8.03: Geometric Progression in Lowest Terms has Coprime Extremes)
 Let $A$, $B$, $C$, $D$ be any multitude whatsoever of numbers in continued proportion (which are) the least of those (numbers) having the same ratio as them.
 I say that the outermost of them, $A$ and $D$, are prime to one another.
 For let the two least (numbers) $E$, $F$ (which are) in the same ratio as $A$, $B$, $C$, $D$ have been taken [Prop. 7.33].
 And the three (least numbers) $G$, $H$, $K$ [Prop. 8.2].
 And (so on), successively increasing by one, until the multitude of (numbers) taken is made equal to the multitude of $A$, $B$, $C$, $D$.
 Let them have been taken, and let them be $L$, $M$, $N$, $O$.
 And since $E$ and $F$ are the least of those (numbers) having the same ratio as them they are prime to one another [Prop. 7.22].
 And since $E$, $F$ have made $G$, $K$, respectively, (by) multiplying themselves [Prop. 8.2 corr.] , and have made $L$, $O$ (by) multiplying $G$, $K$, respectively, $G$, $K$ and $L$, $O$ are thus also prime to one another [Prop. 7.27].
 And since $A$, $B$, $C$, $D$ are the least of those (numbers) having the same ratio as them, and $L$, $M$, $N$, $O$ are also the least (of those numbers having the same ratio as them), being in the same ratio as $A$, $B$, $C$, $D$, and the multitude of $A$, $B$, $C$, $D$ is equal to the multitude of $L$, $M$, $N$, $O$, thus $A$, $B$, $C$, $D$ are equal to $L$, $M$, $N$, $O$, respectively.
 Thus, $A$ is equal to $L$, and $D$ to $O$.
 And $L$ and $O$ are prime to one another.
 Thus, $A$ and $D$ are also prime to one another.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"