$k\geq 2$, there exists integers $r_k,s_k,\ell'_k\geq 1$ such that $\ell_k=\ell'_kp_0^{r_k}p_1^{s_k}$ and $\ell'_k$ is not divisible by $p_0$ or $p_1$ (this is just the decomposition of an integer number in prime factors). Since $\gcd(p_0,p'_2)$ divides all weights, it must be $1$. Hence $p'_2$ is coprime to $p_0$, and similarly is coprime to $p_1$. It follows that $\gcd(\ell_k,\ell_0)=p_1\gcd(\ell'_k,p'_2)$ and that $$\ell_{k:0}=\frac{\ell'_k}{\gcd(\ell'_k,p'_2)}\cdot p_0^{r_k}p_1^{s_k-1} \;.$$ By hypothesis $\ell_{k:0}$ divides $\ell_1$, hence $p_0^{r_k-1}$ and $p_1^{s_k-1}$ divides $p'_2$. Since $p_0,p_1,p'_2$ are pairwise coprime, this implies $r_k=s_k=1$.

Since again $\ell_{k:0}=p_0\ell'_k/\gcd(\ell'_k,p'_2)$ divides $\ell_1$, then $\ell'_k/\gcd(\ell'_k,p'_2)$ divides $p'_2$, which is only possible if $p'_2$ is a multiple of $\ell'_k$. Since $p'_2$ is coprime to $p_0$ and $p_1$, $\ell'_k$ is coprime to $p_0$ and $p_1$ for all $k\geq 2$. Furthermore, $\gcd(\ell'_2,\ldots,\ell'_n)$ divides $\ell_i$ for all $i=0,\ldots,n$, but ${\underline}{\ell}$ is coprime, so ${\underline}{\ell}'$ must be coprime too.

The next step is to show that the weight vector ${\underline}{\ell}'$ satisfies the condition of Lemma \[lemma:divide\]. Let $i\neq j$ and $i,j\geq 2$. Being $\ell'_i$ coprime to $p_0$ and $p_1$, so is $\ell'_{i:j}$. But $\ell'_{i:j}=\ell'_i/\gcd(\ell'_i,\ell'_j)=\ell_{i:j}$ and $\ell_{i:j}$ divides $\ell_k$; hence $\ell'_{i:j}$ divides $\ell'_k$ (for all $k\neq j$, $k\geq 2$).

Now we use the inductive hypothesis, ${\underline}{\ell}'=({\underline}{p}^{\geq 2})^{\,\sharp}$ for some ${\underline}{p}^{\geq 2}=(p_2,p_3,\ldots,p_n)$ pairwise coprime. Since 