 **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. 



A property of periodic words
As you observe, we may assume that $p'$ is empty, i.e., that $v$ starts with a power of $u$. If $p$ is not a multiple of $u$, then writing $u$ as a circular word, it means you can find two district places in the circle where you can read $u$. So $u=xy=yx$ for some $x,y$ (namely if $|p|$ is $r$ mod $|u|$ take $x$ to be the suffix of $p$ of length $r$ and $y$ the prefix of $v$ of length $|u|-r$). But then x,y are powers of a common element and hence u is not primitive. 


