The order of a zero is the exponent in its local holomorphic factorization

For the finite-order-to-factorization direction, suppose ${\mathrm{ord}}_a(f)=m<+\mathrm{\infty }$ and write the Taylor expansion from [L2] as $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$; [L1] gives $c_0=\cdots =c_{m-1}=0$ and $c_m\mathrm{\ne }0$, so formally $f(z)=(z-a{)}^m{\sum }_{k\ge 0}{c}_{m+k}(z-a{)}^k$.

For the factorization-to-finite-order direction, suppose $f(z)=(z-a{)}^{m}g(z)$ locally with $g$ holomorphic and $g(a)\mathrm{\ne }0$; expanding $g(z)={\sum }_{k\ge 0}{b}_k(z-a{)}^k$ by [L2] gives $b_0=g(a)\mathrm{\ne }0$. Multiplication by $(z-a{)}^m$ gives a convergent power-series representation of $f$ whose coefficients below degree $m$ vanish and whose degree-$m$ coefficient is $b_0$; [L5] identifies these with the Taylor coefficients of $f$, so [L1] gives ${\mathrm{ord}}_a(f)=m$.

For the finite-order-to-factorization direction, define $g(z)={\sum }_{k\ge 0}{c}_{m+k}(z-a{)}^k$ on the Taylor disc: at $z=a$ the series has value $c_m$, and away from $a$ its absolute convergence follows by dividing the absolutely convergent tail of the series in step 1.1 by $\mathrm{\mid }z-a{\mathrm{\mid }}^m$; thus $g$ is analytic and [L3] makes it holomorphic.

For the finite-order-to-factorization direction, step 2.1 gives $g(a)=c_m\mathrm{\ne }0$, and [L4] supplies a smaller neighbourhood on which $g$ remains nonzero; hence the required local factorization holds, including $m=0$ where the factor is $1$.

For the infinite-order equivalence, [L1] says infinite order means that every Taylor coefficient is zero, and [L2] then makes $f$ vanish on a neighbourhood of $a$; conversely, if $f$ vanishes on a neighbourhood, all of its derivatives and hence all of its Taylor coefficients at $a$ are zero, so [L1] gives infinite order.