A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by:

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

In other words, it is sufficient that all factorizations of an element with the same number of factors are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the defintion.

Side note: If one replaces 'domain' by 'commutative cancellative semigroup with identity' the weakened definition actually is different.

A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by:

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

In other words, it is sufficient that all factorizations of an element with the same number of factors are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the definition.

Side note: If one replaces 'domain' by 'commutative cancellative semigroup with identity' the weakened definition actually is different.

A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by:

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

In other words, it is sufficient that all factorizations of an element with the \emph{same number of factors} are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the defintion.

Side note: If one replaces domain' by commutative cancellative semigroup with identity' the weakened definition actually is different.

A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by:

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

In other words, it is sufficient that all factorizations of an element with the same number of factors are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the defintion.

Side note: If one replaces 'domain' by 'commutative cancellative semigroup with identity' the weakened definition actually is different.