The following are from Froberg's "Introduction to Grobner bases" and Leo's proof from a different post.
Background
Definition: (Squarefree monomial) A monomial $m$ is called squarefree if $m=x_{i_1},\ldots, x_{i_k}, i_1<\cdots <i_k$. A monomial ideal generated by squarefree momomial is called a squarefree monomial ideal.
Proposition Let $\mathfrak{a}=\langle m_1,\ldots, m_s \rangle$ be a monomial ideal with $m_i=x_{i_1}^{c_{i_1}}\cdots x_{i_k}^{c_{i_k}}, i=1,2,\ldots,s$ and all $c_{i_k}\neq 0$. Then $\sqrt{\mathfrak{a}}=\langle n_1,\ldots, n_s \rangle$, where $n_i=x_{i_1}\cdot\ldots \cdot x_{i_k},i=1,2,\ldots,s$.
Proof: Let $c_i=\text{max }=\{c_{i_k}\mid k=1,2,\ldots,k_i\}$. Then $n_i^{c_i}\in \mathfrak{a}$, and hence we have $\langle n_1,\ldots, n_s \rangle^N\subseteq \mathfrak{a}$ with $N=c_1+\cdots +c_s - s+1$ since $(f_1n_1+\cdots+f_sn_s)^N$ is a sum of of terms $g_in_1^{d_1}\cdots n_s^{d_s}$ with $d_1+\cdots +d_s=N$, so $d_i\geq c_i$ for at least one $i$, therefore every term belongs to $\mathfrak{a}$. Since
$$\langle n_1,\ldots, n_s \rangle^N\subseteq \mathfrak{a}\subseteq \langle n_1,\ldots, n_s \rangle$$
we get
$$\sqrt{\langle n_1,\ldots, n_s \rangle^N}\subseteq \sqrt{\mathfrak{a}}\subseteq \sqrt{\langle n_1,\ldots, n_s \rangle},$$
But
$$\sqrt{\langle n_1,\ldots, n_s \rangle^N}=\sqrt{\langle n_1,\ldots, n_s \rangle}=\langle n_1,\ldots, n_s \rangle$$
since $\langle n_1,\ldots, n_s \rangle$ is squarefree.
Questions
The proof for the above Proposition which I have questions about has already been given in this post in @Leo's answer. What I have trouble understanding from that post and also from the above proof are as follows:
$1$ Where does the number $N=c_1+\cdots +c_s - s+1$ come from?
$2.$ why does $N$ also equal to $d_1+\cdots +d_s$? I mean how are the two different values of $N$ equal.
[Instead of $N$, it is denoted as $k$ in Leo's answer]
Also,
$3.$ when in Froberg's proof above, when it says: "$g_in_1^{d_1}\cdots n_s^{d_s}$", should it not be $f_in_1^{d_1}\cdots n_s^{d_s}$ instead?
For convenience, here is Leo's answer:
Claim: If $m_1,\ldots,m_s$ are monomials in $K[x_1,\ldots,x_n]$, then $$\sqrt{\langle m_1,\ldots,m_s\rangle} = \langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle.$$
Proof: Put $k_i:=(\text{greatest exponent of any variable in }m_i)\in\mathbb{N}$, i.e. if $m_i=x_{j_1}^{a_1}\cdots x_{j_l}^{a_l}$ then $k_i=\max\{a_1,\ldots,a_l\}$. Now put $k:=k_1+\cdots+k_s-s+1$. We have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq \langle m_1,\ldots,m_s\rangle$, because every term of $\prod_{j=1}^k(f_{1,j}\sqrt{m_1}+\cdots+f_{s,j}\sqrt{m_s})$ has the form $f\sqrt{m_1}^{\beta_1}\cdots\sqrt{m_s}^{\beta_s}$ where $b_j\in\mathbb{N}_0$ and $\beta_1+\cdots+\beta_s=k$, which means at least one $\beta_j\geq k_j$. Therefore $$\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle^k\subseteq\langle m_1,\ldots,m_s\rangle\subseteq\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle/\sqrt{~},$$ $$\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}\subseteq\sqrt{\langle m_1,\ldots,m_s\rangle}\subseteq\sqrt{\langle\sqrt{m_1},\ldots,\sqrt{m_s}\rangle}.$$ Thus it remains to show that $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle}=\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle$, i.e. that squarefree monomial ideals are radical.
If $\sqrt{m_1}=x_{j_1}\cdots x_{j_l}$, we have $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_{r=1}^l\langle x_{j_r},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle$, because by point e) from my post here, $$\langle x_{j_1},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle\cap\langle x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s}\rangle= \sum\sum\langle\mathrm{lcm}(\ast,\ast)\rangle= \langle x_{j_1}x_{j_2},\sqrt{m_2},\ldots,\sqrt{m_s} \rangle.$$ Next, $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle=\bigcap_r\bigcap_{r'}\langle x_{j_r},x_{j_{r'}},\sqrt{m_3},\ldots,\sqrt{m_s}\rangle$, and so on. Therefore $\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle = \bigcap_\lambda\mathfrak{p}_\lambda$ for some ideals $\mathfrak{p}_\lambda$, generated by variables. But $\mathfrak{p}_\lambda$ are prime by b), hence $\sqrt{\langle \sqrt{m_1},\ldots,\sqrt{m_s}\rangle} = \sqrt{\cap_\lambda\mathfrak{p}_\lambda}=\cap_\lambda\sqrt{\mathfrak{p}_\lambda}=\cap_\lambda\mathfrak{p}_\lambda$, since prime ideals are radical. $\blacksquare$
Thank you in advance.
