I am reading the book linear algebraic groups by Springer. I have a question on Page 53, on line 3, it is said that $d-h \geq p$ implies that ${d-h \choose p} \not\equiv 0 \pmod p$ by Lemma 3.4.2. But it seems that this is not true. For example, let $d=p^2+p, h=p-1$. Then $p$ divides $d$ and does not divide $h$. $p \leq d-h$ but ${d-h \choose p} \equiv 0 \pmod p$. Thank you very much.
