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. $\blacksquare$
One can show that if $f_1, \ldots, f_n, g$ are monomials, then $g \in \langle f_1, \ldots, f_n \rangle$ if and only if some $f_k \mid g$.
This is theorem $1.1.8$ of Moore, Rogers, and Sather-Wagstaff's Monomial Ideals and their Decompositions, available here.
Now if $f_1 = gh$, then $g$ and $h$ must be monomials (this is intuitively clear, but it's also proven in Chapter $1$ of the same book). So if $g$ were in $\langle f_1, \ldots, f_n \rangle$, we would need some $f_k \mid g$. But then we would have $f_k \mid f_1$, so $f_1$ would be redundant, contradicting minimality.
I hope this helps ^_^
Best Answer
Take a minimal system of (monomial) generators for $I$. If $x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ with $q_i\ge 1$ is one of them, let's suppose that some $q_t\ge 2$. Then $x_{i_1}\cdots x_{i_r}\in\sqrt I$, and $x_{i_1}\cdots x_{i_r}\notin I$.
If $x_{i_1}\cdots x_{i_r}\in I$, then there is a monomial $m$ in the minimal system of generators of $I$ such that $m\mid x_{i_1}\cdots x_{i_r}$. Since $x_{i_1}\cdots x_{i_r}\mid x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$ we must have $m=x_{i_1}^{q_1}\cdots x_{i_r}^{q_r}$, but this is not possible because $x_{i_t}$ appears in $m$ at most once.