You acknowledge in a comment that you just want to construct $x^n$. Well, given a segment $\overline{OP_0}$ of length $1$, and a segment $\overline{OP_1}$ of length $p$, then one can construct segments $\overline{OP_n}$ of length $p^n$ with a simple spiral of perpendicular lines:
The construction leverages the proportions of the similar right triangles $\triangle OP_nP_{n-1}$ and $\triangle OP_{n-1}P_{n-2}$ for which we have
$$\frac{|\overline{OP_n}|}{|\overline{OP_{n-1}|}} = \frac{|\overline{OP_{n-1}}|}{|OP_{n-2}|} \qquad\to\qquad |\overline{OP_n}| = \frac{\;|\overline{OP_{n-1}}|^2}{|OP_{n-2}|}$$
If the "earlier" segments obey the relation $|\overline{OP_k}| = p^k$ (which they certainly do for $k = 0$ and $k=1$), then we have
$$|\overline{OP_n}| = \frac{\left(p^{n-1}\right)^2}{p^{n-2}} = \frac{p^{2n-2}}{p^{n-2}} = p^n$$
so that segment $\overline{OP_n}$ satisfies the relation, too. By induction, we get any (non-negative integer) power of $p$ we like. (Note that we can continue the spiral "backwards" from $P_0$ to get the negative-integer powers $p^{-1}$, $p^{-2}$, $\dots$ .)
Observe that having an auxiliary segment of length $1$, against which to compare the segment of length $p$, is crucial to this construction ... and to any construction, really. If we have a segment of length $p$, and nothing else, then there's no telling what a segment of length $p^2$ would look like: If $p$, compared to some "hidden" unit, turns out to be, say, $2$, or $1/2$, or $1$, then the segment of length $p^2$ would either be twice as long ($p^2 = 4 = 2p$), half as long ($p^2 = 1/4 = p/2$), or just as long ($p^2 = 1 = p$) as the original segment. (We don't have this problem when turning a segment of length $p$ into a square of area $p^2$, because any "hidden" unit of length is automagically compatible with the corresponding "hidden" unit of area to make this work.)
I don't know that this helps with understanding the Extended Pappus Four-Line Problem, however.
