This concerns the paragraph just before equation 2.9. I will quote it here

The sum of the geometric series $2^{-1} + 2^{-2}+ \cdot\cdot\cdot + 2^{-n} = (\frac{1}{2})^1 + (\frac{1}{2})^2 + \cdot\cdot\cdot + (\frac{1}{2})^n$ will be derived later in this chapter; it turns out to be $1 - (\frac{1}{2})^n$. Hence $T_n = 2^nS_n = 2^n - 1$

That last bit

\[T_n = 2^nS_n = 2^n - 1\]

Had me confused for a while but, after stopping overthinking it, I came to release they mean things quite literally using the previous definitions in the book with some algebraic manipulations

\[\begin{align} \frac{T_n}{2^n} &= S_n \\ T_n &= 2^nS_n \\ \end{align}\]

Now at this point we can substitute in the definition for $S_n$ which is the sum of the geometric series given to us in the quote above as $1 - (\frac{1}{2})^n$. First we slightly rearrange it as $1 - \frac{1}{2^n}$ then plug that in

\[\begin{align} T_n &= 2^n(1 - \frac{1}{2^n}) \\ T_n &= 2^n - \frac{2^n}{2^n} \\ T_n &= 2^n - 1 \end{align}\]

Tada! We get the closed form for the Towers of Hanoi recurrence.