** Question is edited** Perhaps this formulation is clearer.

It is well known that if a power of a primitive (i.e. not a proper power) word $u$ contains two different occurrences of a word $v$, $|v|>|u|$, then the occurrences are shifts of each other by a multiple of $|u|$ (formally: $u^n\equiv pvq\equiv p'vq', |p|-|p'|\in |u|\mathbb{Z}$). What is the simplest proof of that fact? Is there a simple proof without using Fine-Wilf?

** Added. A sketch of a proof.** We can assume that $v$
starts with $u$, $v\equiv u^sw$ for $s\ge 1$. Then $v\equiv u^sw\equiv
u_1u^sw_1$ such that $|u_1|>0$, $|u_1|+|w_1|=|w|$. Therefore $w\equiv w_2w_1$
for some $w_2$ whence $u^sw_2\equiv u_1u^s$. Then a standard fact from "combinatorics on words" gives
that $u_1\equiv xy, w_2\equiv yx, u^s\equiv (xy)^mx$ for some words $x,y$ and some $m\ge 1$ (the fact is: if $pq=qr$, then $p\equiv p_1p_2, q\equiv (p_1p_2)^zp_1, q\equiv p_2p_1$ for some $p_1,p_2,z$). Therefore $u_1u^s\equiv xy(xy)^mx$ is periodic with periods $u$ and $xy$. Its length is at least $|xy|+|u|$. Therefore by Fine-Wilf it is periodic with period $t$ such that $|t|$ divides $|xy|=|u_1|$ and $|u|$. Hence $u$ is a power of a proper subword of $u$, a contradiction.