YaoAOPS wrote:

Very cool problem.

We instead consider the expression $1 + \frac{(1-p)x}{1 + \frac{(2-p)x}{1 + \frac{\ddots}{1 - x}}} = \frac{P_{p-1}(x)}{Q_{p-1}(x)}$ We then have the relation $\frac{P_k(x)}{Q_k(x)} = 1 + \frac{-kx}{\frac{P_{k-1}(x)}{Q_{k-1(x)}}} = \frac{P_{k-1}(x) -kx Q_{k-1}(x)}{P_{k-1}(x)}$ We inductively then have that $\gcd(P_i, Q_i) = 1$ . Define $P_1' = 1-x, P_0' = 1$ , so this rewrites as $P_k'(x) = P_{k-1}' - kx P_{k-2}'(x)$ .
Now, replace $x$ with $-x$ and shift to instead get $P_0 = 1, P_1 = 1, P_k = P_{k-1} + (k-1)x P_{k-2}$ , and we want to show that instead $P_{p} \equiv 1 \pmod{p}$ and $P_{p-1}$ has all nonzero coefficients.
Here's a table of the first few entries of $p$ .
$\begin{tabular}{c|ccccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ $x$ & 0 & 0 & 1 & 3 & 6 & 10 & 15 & 21 & 28 \\ $x^2$ & 0 & 0 & 0 & 0 & 3 & 15 & 45 & 105 & 210 \\ $x^3$ & 0 & 0 & 0 & 0 & 0 & 0 & 15 & 105 & 420 \\ $x^4$ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 105 \end{tabular}$
Claim: We have that $[x^k]P_x = A_k(x)$ where $A_k(x) = \frac{x(x-1) \cdots (x-(2k-2))(x-(2k-1))}{2^k k!}$ Proof. We have that $[x^t] P_k = [x^t] P_{k-1} + [x^{t-1}] (k-1) P_{k-2}$ . This rewrites as showing that $A_t(k) - A_t(k-1) = (k-1) A_{t-1}(k-2)$ Expand this and factor out $\frac{1}{2^{t-1} (t-1)!} (k-2) \cdots (k-(2t+1))$ to simplify this as $\frac{1}{2t} \left((k(k-1) - (k-1)(k - 2t))\right) = k-1$ which finishes. $\blacksquare$

This then finishes by plugging in $t = p, t = p-1$ .

Remark: Trying to represent each row as a polynomial follows from noticing the fact that the first three rows are in fact $1$ , $\binom{n}{2}$ , and $\binom{\binom{n}{2}}{2}$ or noticing that $P_k \equiv P_{p} \pmod{p}$ .
Getting this specific form for the polynomial comes from fundamental theorem of algebra.
As a matter of fact, this is enough to prove that $P_{p-1} \equiv 1 \pmod{p}$ .

This is also the initial solution I found while testsolving. I however don't see a way to see how to show the original $P,Q$ are coprime using this method, since the image of $P$ in $\mathbb{F}_p[x]$ is just 1. This is very minor.