Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 3 \pmod 4} $$
Lemma: $$ (-p|q) = (q|p). $$
Lemma: If $$ a^2 + p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
Let $$ F_1 = 4 + p, $$
$$ F_2 = 4 F_1^2 + p, $$
$$ F_3 = 4 F_1^2 F_2^2 + p, $$
$$ F_4 = 4 F_1^2 F_2^2 F_3^2 + p, $$
$$ F_5 = 4 F_1^2 F_2^2 F_3^2 F_4^2 + p, $$
and so on.
These are all of the form $a^2 + p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
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Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 1 \pmod 4} $$
Lemma: $$ (p|q) = (q|p). $$
Lemma: If $$ a^2 - p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
FIND an even square $$ W = 4^k = \left( 2^k \right)^2 $$ such that
$$ \color{magenta}{ W > p.} $$
Let $$ F_1 = W - p, $$
$$ F_2 = W F_1^2 - p, $$
$$ F_3 = W F_1^2 F_2^2 - p, $$
$$ F_4 = W F_1^2 F_2^2 F_3^2 - p, $$
$$ F_5 = W F_1^2 F_2^2 F_3^2 F_4^2 - p, $$
and so on. As $p \equiv 1 \pmod 4$ and $W \equiv 0 \pmod 4,$ we know $W - p \equiv 3 \pmod 4 $ and so $W-p \geq 3. $ So the $F_j$ are larger than $1$ and strictly increasing.
These are all of the form $a^2 - p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
Without Euclid's Lemma or Bezout's Identity & all that.
Preface. We consider "$1$" to be the unique prime factorization of $1$, so as to not need to discuss some special cases separately.
By contradiction, suppose $\emptyset \ne E= \{A\in \Bbb N : A \text { has more than one prime factorization }\}.$ Let $P =\min E.$ Then for some unequal increasing finite sequences $(p_1,...,p_m)$ and $(q_1,...,q_n)$ of primes we have $$(1)...\quad P=\prod_{i=1}^mp_i=\prod_{j=1}^nq_j.$$
We have $m\ge 2$ and $n\ge 2,$ otherwise the prime $p_1=P$ is divisible by some prime $q_j\ne p_1,$ or the prime $q_1=P$ is divisible by some prime $p_i\ne q_1.$
Now if any $p_i=q_j=r$ for any $i,j$ then we could divide $(1)$ by $r$ and get $\min E=P>P/r\in E,$ which is absurd. So no $p_i$ equals any $q_j.$
Let $\prod_{i=2}^mp_i=X$ and $\prod_{j=2}^n q_j=Y.$ So we can write $$P=p_1X=q_1Y. $$ WLOG (without loss of generality) let $q_1<p_1.$ There exists a (unique) $k\in \Bbb N$ such that $kq_1\le p_1<(k+1)q_1.$ Let $s=p_1-kq_1. $ We have $s\ne 0$ otherwise the prime $p_1$ would be divisible by the smaller prime $q_1$.
So $1\le s=p_1-kq_1<(k+1)q_1-kq_1=q_1<p_1.$
We have $$0<X\le sX=(p_1-kq_1)X=p_1X-kq_1X=$$ $$=P-kq_1X=$$ $$=q_1Y-kq_1X=q_1(Y-kX).$$ $(2).$ So we have $0<sX=q_1(Y-k X).$ This also implies $Y-kX\ge 1.$
Now $X$ has a prime factorization that does not include $q_1$ and no prime factorization of $s$ can include $q_1$ because $s<q_1.$
$(3).$ So $sX$ has a prime factorization that $does$ $not$ include $q_1.$
$(4).$ And $q_1$ times any prime factorization of $(Y-kX)$ is a prime factorization of $q_1(Y -kX)$ that $does$ include $q_1.$
But $(2).sX=q_1(Y-kX).$ So by $(3)$ and $(4)$ we have $sX\in E.$ This is absurd because $sX<p_1X=P=\min E.$
Best Answer
First, we note that $p|q_1$ implies $p=q$ (we don't even need Euclid's lemma to see this).
Now assume that $p|q_1q_2\ldots q_k\implies p|q_1$ or $p|q_2$ or $\cdots$ or $p|q_k$.
Now consider the statement $p|q_1q_2\ldots q_kq_{k+1}$. Using Euclid's lemma, $p$ must divide one of $q_1q_2\ldots q_k$ or $q_{k+1}$. By our inductive hypothesis, we conclude that $p|q_1$ or $p|q_2$ or $\cdots$ or $p|q_k$ or $p|q_{k+1}$.
I see that this is pretty much what you did, so which part specifically are you having difficulty writing?