The ring of holomorphic germs is a UFD

The proof is by induction on $m$. For $m=1$, every nonzero nonunit germ has the form $z_1^{d}u$ with $d\ge 1$ and $u$ a unit by [L5]. Thus the only irreducible germs are the associates of $z_1$, and every factorization is determined uniquely by the zero order. So ${\mathcal{O}}_{1,0}$ is a UFD.

Assume $m>1$ and that $R:={\mathcal{O}}_{m-1,0}$ is a UFD. Then $R$ is a domain by [L1], so its field of fractions $K=\mathrm{Frac}(R)$ exists by [L2], and $K[z_m]$ is a UFD by [L2]. Using [L3], every primitive polynomial in $R[z_m]$ is irreducible there exactly when it is irreducible in $K[z_m]$, and products of primitive polynomials remain primitive. Therefore every nonzero polynomial in $R[z_m]$ factors uniquely, up to order and associates, by first factoring in $K[z_m]$ and then clearing denominators. Hence $R[z_m]$ is a UFD.

Let $f\in {\mathcal{O}}_{m,0}$ be a nonzero nonunit. By [L5], after a complex-linear coordinate change $T$ the pulled-back germ $T^{\ast }f=f\circ T$ is regular in $z_m$. By [L4], write $T^{\ast }f=uW$ with $u$ a unit and $W\in R[z_m]$ a Weierstrass polynomial. Since $R[z_m]$ is a UFD by step 1.2, factor $W=P_1\cdots P_s$ into irreducible polynomials. The correspondence in [L4] turns this into an irreducible factorization of $T^{\ast }f$, and applying $T^{-1}$ gives an irreducible factorization of $f$.

For uniqueness, let $f=q_1\cdots q_t$ be any factorization of $f$ into irreducible germs. Applying $T^{\ast }$ gives a factorization of $T^{\ast }f$. Since $T^{\ast }f$ is regular, [L4] makes each $T^{\ast }{q}_j$ regular and gives prepared polynomials $Q_j\in R[z_m]$ whose product is $W$. Step 1.2 gives uniqueness of the factorization of $W$ in $R[z_m]$, so after reordering each $Q_j$ is associate to one of the $P_i$. Then [L4] makes the corresponding germs $T^{\ast }{q}_j$ associate to the prepared factor coming from $P_i$, and applying $T^{-1}$ returns uniqueness for the original factorization of $f$.

It remains to check that ${\mathcal{O}}_{m,0}$ is a domain. Suppose $ab=0$ as germs on a connected polydisc. If $a$ were nonzero, the set where $a\mathrm{\ne }0$ would be a nonempty open subset, and on it $b=0$; [L7] would force $b=0$ on the whole polydisc. Thus $ab=0$ implies $a=0$ or $b=0$. For $m>1$, steps 2.1, 3.1, and 4.1 therefore give existence, uniqueness, and the domain property required by [L1]; together with the base case in step 1.1, this completes the induction.