If $1$ were admitted as a prime, uniqueness would fail: $6 = 2 \cdot 3 = 1 \cdot 2 \cdot 3 = 1 \cdot 1 \cdot 2 \cdot 3$, lists of different lengths that no permutation matches

Refuted claim. The clause $p > 1$ in Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ is an arbitrary convention: replacing it by $p \ge 1$ would leave The fundamental theorem of arithmetic: every integer $n \ge 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i<r} p_i = \prod_{j<s} q_j$ with every $p_i$ and $q_j$ prime, then $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$ true as stated.

Write prime$^{*}$ for the modified notion — $p \ge 1$ and the only positive divisors of $p$ are $1$ and $p$ — so that $1$ is prime$^{*}$ and every prime is prime$^{*}$. The claim under refutation is that for lists of prime$^{*}$s, $\prod_{i<r} p_i = \prod_{j<s} q_j$ still forces $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$ (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition, The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity).

Witness. The lists $(2,3)$, $(1,2,3)$ and $(1,1,2,3)$, of lengths $2$, $3$ and $4$, all consist of prime$^{*}$s and all have product $6$. So $r = s$ fails already between the first two, and no permutation can exist because the index sets have different sizes.

Numerals. For $k \in \mathbb{N}$ the symbol $k$ inside $\mathbb{Z}$ means $\iota(k)$, the embedding of The naturals embed in the integers.

• The failure is not a technicality about lists. Under the modified definition an integer would have infinitely many factorisations, one for each number of padding factors $1$, so no formulation of uniqueness survives: neither "the same length", nor "the same multiset of factors", nor "the same exponent vector", since $1$ would carry an arbitrary exponent.

• Existence is what makes the convention a genuine choice rather than a necessity. Both notions give factorisations of every $n \ge 1$, and the difference shows up only when one asks whether the factorisation is unique. That is exactly the asymmetry Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ appeals to when it excludes $1$.

• $0$ is excluded for a different reason. It is not that $0$ would break uniqueness; it is that $0$ has every positive integer as a divisor, so it fails the divisor clause outright, and a product containing the factor $0$ is $0$ rather than the integer being factored.