#### Answer

The proof is below.

#### Work Step by Step

We know the following:
$dQ = nC_pdT$
We also can take the derivative of both sides of the equation $ \Delta S = \frac{ Q}{T}$ to find :
$ d S = \frac{ dQ}{T}$
Combining these two equations, it follows:
$ dS = \frac{ nC_pdT}{T}$
$ S = \int_{T_1}^{T_2}\frac{ nC_pdT}{T}$
$S = nC_p ln (\frac{T_2}{T_1})$