To find the powers, you must iterate, but there are efficient ways of doing it.
For starters, each time you find a prime factor $p$, factor it out of $n$ before looking for new factors. For your example:
$$567788=2*283894=2*2*141947$$
We factor $2$ out until it doesn't come out anymore, then look for the next factor. 3,5,7,11 are not factors, 13 is, so
$$=2*2*13*10919$$
13 does not divide again, keep trying factors until finding 61 works
$$=2*2*13*61*179$$
We know that 179 is prime without having to try any other prime factors because we already know it cannot be divisible by anything less than the last prime factor we found (61) which is more than $\sqrt{179}$. Notice we did not need to try dividing by any number larger than 61, which is a good deal less than $\sqrt{567788}\approx750$.
There are certainly other ways of doing this but this is both simple to understand and implement and saves you a lot of computation from your previous method.
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
Currently very little is known about this problem and it appears intractable by known methods, though it is of great interest. More generally, additive number theory takes upon the challenge of studying the additive structure of prime numbers, which is bound to be difficult due to their inherent multiplicative nature.
Some problems that would greatly benefit from knowing how addition effects prime factorizations include: The Twin Prime Conjecture and The Collatz Conjecture.