Simple-root lifting and factor lifting produce the same root

By [L1], the complete local ring $k\mathrm{⟦}t\mathrm{⟧}$ is Henselian. Modulo $t$, one has $\bar{f}=T(T-1)$, so $\bar{a}=1$ is a simple root. By [L2], there is a unique root $a\in k\mathrm{⟦}t\mathrm{⟧}$ with $a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}t)$ and $f(a)=0$. The first two correction stages give $a\equiv 1+t(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2),a\equiv 1+t-t^2(\mathrm{m}\mathrm{o}\mathrm{d}{t}^3)\mathrm{.}$

The residue factorization $\bar{f}=T(T-1)$ is monic and coprime, so [L3] gives a factorization $f=(T-b)h(T)$ with $b\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}t)$. Evaluating at $T=b$ gives $f(b)=0$, so $b$ is another lift of the same simple residue root.

By uniqueness in [L2], one has $a=b$. Equivalently, the root obtained from the lifted linear factor is exactly the same root obtained from the simple-root lifting procedure.