Below is one simple proof (which has the nice property that it is conceptually faithful to the essence of the matter). I've had much success explaining this proof to educated laypersons, bright high-school students (and even some talented elementary students). Since you say you already know a few proofs, I will omit the easy parts, such as the simple inductive proof of the existence of prime factorizations. Instead, I concentrate only on the most difficult step: the key inductive step of the uniqueness of prime factorizations, viz. Euclid's Lemma. That follows immediately from the following fundamental lemma - which has widespread applications in number theory.
Lemma $\ $ If $\rm\,S\,$ is a nonempty set of positive integers that is closed under positive subtraction (i.e. for any $\rm\:m,n\in S,\:$ if $\rm\: m > n\:$ then $\rm\: m-n\in S),\:$ then every integer in $\rm\:S\:$ is a multiple of the least integer $\rm\:d\:$ in $\rm\:S.$
Proof $\ $ Suppose $\rm\:s\in S.\:$ By repeatedly subtracting $\rm\:d,\:$ we deduce that $\rm\:S\:$ contains all of the positive integers in the sequence $\rm\ s,\ s\!-\!d,\ s\!-\!2d,\ s\!-\!3d,\ldots$ By the Division Algorithm, the least nonegative integer in this sequence is the remainder $\rm\:r = s-qd\:$ left on dividing $\rm\:s\:$ by $\rm\:d,\:$ and $\rm\:0 \le r < d.\:$ The remainder $\rm\:r\:$ must be $\,0\:$ $\rm(\Rightarrow\:s = qd),\:$ else $\rm\:0 < r< d,\:$ contra $\rm\:d\:$ is the least element of $\rm\:S.\ $ QED
For example, the positive even integers $\rm\:S = \{2,4,6,\ldots\}\,$ are closed under positive subtraction since $\rm\: 2m-2n = 2(m-n)\:$ is even. Thus the Lemma implies that every positive even integer is a multiple of the least positive even integer, i.e. $\rm\:2n\:$ is a multiple of $\,2.\:$ Similarly for $\rm\:S = \{3,6,9\},\:$ but not for $\rm\:\{6,9\}\:$ since it is not closed under positive subtraction: $\rm\:9\!-\!6=3\not\in S.\:$ One corollary of the Lemma is the following result, which is the keystone of uniqueness of prime factorizations.
Euclid's Lemma $\ \ $ Prime $\rm\ p\mid ab\:\Rightarrow\: p\mid a\ \ or\ \ p\mid b, \ $ for all positive integers $\rm\ a,b.$
Proof $\ $ Let $\rm\:S\:$ be the set of positive integers $\rm\:n\:$ such the $\rm\:p\mid an,\:$ i.e. $\rm\:p\:$ divides $\rm\:an.\:$ Note $\rm\:S\:$ is nonempty since $\rm\:p\mid ap,ab\:\Rightarrow\:p,b\in S.\:$ Also $\rm\:S\:$ is closed under positive subtraction, since if $\rm\:m,n\in S\:$ and $\rm\:m>n\:$ then $\rm\:p\mid am,an\:\Rightarrow\:p\mid am\!-\!an = a(m\!-\!n),\:$ so $\rm\:m\!-\!n\in S.\:$ Let $\rm\,d\,$ be the least element of $\rm\:S.\:$ By the Lemma, $\rm\:d\:$ divides every element of $\rm\:S.\:$ Thus $\rm\:d\mid p.\:$ But the only divisors of the prime $\rm\:p\:$ are $\rm\:d = 1\:$ or $\rm\:d = p.\:$ In the first case, $\rm\:d = 1\in S,\:$ so $\rm\:p\mid a\cdot 1 = a,\:$ by definition of $\rm\:S.\:$ In the second case, $\rm\:d = p\:$ so, by the Lemma, $\rm\:p\mid b,\:$ by $\rm\:b\in S.\:$ Thus either $\rm\:p\mid a\:$ or $\rm\:p\mid b.\ \ $ QED
Now it is straightforward to inductively apply Euclid's Lemma to match up, then cancel primes in any two prime factorizations of an integer, so inductively proving the uniqueness of prime factorizations, e.g. see this proof, which uses a fractional version of the Lemma. That page also includes many other proofs you may find of interest, including a famous proof of Zermelo, which eliminates the division algorithm by directly inlining into the proof its inductive essence. One can also do the same for the above Lemma if one so desires, e.g. see here. But that only serves to obscure the conceptual essence of the matter (the Euclidean algorithm, or principality of ideals, etc). The point of abstracting out the above lemma is that this (ideal) structure is ubiquitous in number theory and algebra - something that will become much clearer when one studies (algebraic) number theory and university algebra.
