Let $R$ be a commutative noetherian ring containing a field of characteristic $p\gt0.$ For an ideal $I\subset R,$ the tight closure $I^*$ is defined as $$\{f\in R\mid \exists t\in R, t\not\in\mathfrak p, \forall \mathfrak p \text{ minimal prime of }R,\ tf^q\in I^{[q]} \text{ for almost all }q=p^e\},$$ where $I^{[q]}$ is the extended ideal of $I$ under the Frobenius homomorphism $R\to R.$ An ideal is said to be tightly closed if it coincides with its tight closure. And my question is
Why is the tight closure of an ideal $I$ tightly closed?
I tried writing out the definitions and see why this holds, but to no avail: there seems to miss something to conclude that $(I^*)^*\subset I^*.$
Any hint is welcomed and thanked, as well as any references.
P.S. I saw this statement in the book Three lectures on commutative algebra.
Take $x_1,\dots,x_n$ a system of generators for $I^*$. By definition there exists $c_i\in R^0$ such that $c_ix_i^q\in I^{[q]}$ for $q\gg0$. Set $c=c_1\cdots c_n$. Then $c(I^*)^{[q]}\subset I^{[q]}$ for $q\gg0$.
Now let $x\in (I^*)^*$. There exists $c'\in R^0$ such that $c'x^q\in (I^*)^{[q]}$ for $q\gg0$. Then $cc'x^q\in I^{[q]}$ for $q\gg0$, so $x\in I^*$.