$\displaystyle\sum_{k=1}^{p-1} \frac 1k = 1+\frac 12 + \frac 13 + \ldots + \frac {1}{p-3}+\frac{1}{p-2}+\frac{1}{p-1}$

$=(1+\frac {1}{p-1})+(\frac 12 +\frac 1{p-2})+(\frac 13 + \frac 1{p-3})+\ldots$

$=\frac{p}{p-1}+\frac{p}{2(p-2)}+\frac{p}{3(p-3)}+\ldots$

All numerators has $p$

Then numerator of the sum will also have p