Given a prime $p$ and a positive integer $n<p$, let $R(p,n)$ be the multiplicative order of $p$ modulo $n!$.
In other words, $R(p,n)$ is the minimal positive integer $r$ such that

$$p^r\equiv 1(\mathrm{mod}n!)$$

For example, $R(7,4)=2$ and $R({10}^9+7,12)=17280$.

Find $R({10}^9+7,{10}^7)$. Give your answer modulo ${10}^9+7$.