I will explain my comment for the case $n=2$ (have not carefully checked the other cases, but I do not see any problems except notational difficulties).

So, assume $\sqrt{I}=I_1I_2$ as above. Pick $x\in I_1, y\in I_2$ such that $x+y=1$. Define $J_1=\{a\in R| y^ma\in I\,\text{for some}\, m\}$ and similarly $J_2$. Since $xy\in \sqrt{I}$, we have $x^Ny^N\in I$ and thus, $x^N\in J_1, y^N\in J_2$, in particular they are comaximal.

Next, we show that $\sqrt{J_1}=I_1$ (and similarly for $I_2,J_2$). If $a\in J_1$, we have $y^ma\in I$. We can write $px+qy^m=1$ for some $p,q\in R$ and thus $a=pxa+qy^ma$. One has $pxa\in I_1$, since $x$ is. $qy^ma\in I\subset I_1$. So, we get $a\in I_1$. Conversely, let $b\in I_1$. Then $by\in I_1I_2$ and thus $b^my^m\in I$ for some $m$ and then, $b^m\in J_1$. 

Finally, we show that $J_1J_2=I$. Let $a\in J_1, b\in J_2$. Then $x^mab\in I$ and $y^nab\in I$ and thus $ab\in I$, since $x^m, y^n$ are comaximal. Conversely, let $a\in I$. Write $px^N+qy^N=1$ for $p,q\in R$. Then, $a=px^Na+qy^Na$. Since $px^Na\in I$, we have $pa\in J_2$. But, $x^N\in J_1$, so $px^Na\in J_1J_2$ and similarly, $qy^Na\in J_1J_2$ and so $a\in J_1J_2$ finishing the proof.