Proof: By Euclid

Let $A$, $B$, $C$ be any multitude whatsoever of numbers in continued proportion, (such that) as $A$ (is) to $B$, so $B$ (is) to $C$.
And let $A$, $B$, $C$ make $D$, $E$, $F$ (by) multiplying themselves, and let them make $G$, $H$, $K$ (by) multiplying $D$, $E$, $F$.
I say that $D$, $E$, $F$ and $G$, $H$, $K$ are in continued proportion.

fig13e

For let $A$ make $L$ (by) multiplying $B$.
And let $A$, $B$ make $M$, $N$, respectively, (by) multiplying $L$.
And, again, let $B$ make $O$ (by) multiplying $C$.
And let $B$, $C$ make $P$, $Q$, respectively, (by) multplying $O$.
So, similarly to the above, we can show that $D$, $L$, $E$ and $G$, $M$, $N$, $H$ are in continued proportion in the ratio of $A$ to $B$, and, further, (that) $E$, $O$, $F$ and $H$, $P$, $Q$, $K$ are in continued proportion in the ratio of $B$ to $C$.
And as $A$ is to $B$, so $B$ (is) to $C$.
And thus $D$, $L$, $E$ are in the same ratio as $E$, $O$, $F$, and, further, $G$, $M$, $N$, $H$ (are in the same ratio) as $H$, $P$, $Q$, $K$.
And the multitude of $D$, $L$, $E$ is equal to the multitude of $E$, $O$, $F$, and that of $G$, $M$, $N$, $H$ to that of $H$, $P$, $Q$, $K$.
Thus, via equality, as $D$ is to $E$, so $E$ (is) to $F$, and as $G$ (is) to $H$, so $H$ (is) to $K$ [Prop. 7.14].
(Which is) the very thing it was required to show.
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