This is the proof I've found for the `Carnot Factor`:

$$S = \frac{Q}{T} \qquad \qquad (1)$$

$$S\_{in} = S\_{out}\qquad \qquad (2)$$

Because entropy stays the same when using an ideal engine:
$$ (1) \& (2) \Rightarrow \frac{Q\_{in}}{T\_{in}} = \frac{Q\_{out}}{T\_{out}} \qquad \qquad (3)$$
$$ (1) \& (2) \Rightarrow Q\_{out}= Q\_{in}\frac{T\_{out}}{T\_{in}} \qquad \qquad (3)$$

From the energy conservation law

$$ W = Q\_{in} - Q\_{out}\qquad \qquad (4) $$

Then from (5) (3) & (4), the Carnot factor $\eta $ is calculated as:

$$ W = Q\_{in} - Q\_{in}\frac{T\_{out}}{T\_{in}}$$
$$ W = Q\_{in}\left(1 - \frac{T\_{out}}{T\_{in}}\right) $$
$$ \eta = \frac{W}{Q \_{in}} = \left(1 - \frac{T\_{out}}{T\_{in}}\right) $$

I don't see, why $T\_{out}$ must be smaller than $T\_{in}$.
There are no restrictions that says $T\_{in} > T\_{out}$. It's just necessary, that $S\_{in} = S\_{out}$. Maybe my book simplifies certain aspects?