Assume there exists a simple group with order $525$
$525 = 3*5*5*7$
we get $n_3 \in \{1, 5, 7, 25, 35, 150\}, n_5 \in \{1, 21\}, n_7 \in \{1, 3, 5, 15\}$
Assuming the group is simple $n_5 = 21, n_7 = 15, n_3 = 7 \, or \, 25$
now $|G| \nmid 8!$ so $n_3 = 25$
now using a lemma we get $21 \not\equiv 1 (mod\,25)$ thus $\exists P,R \in Syl_5(G)$ such that $P \cap R \neq \{e\}, P \neq R$ and $|P \cap R| \mid P , R$ so $|P,R : P \cap R| = 5 $ and $| P \cap R | = 5$
Consider $N_G(P\cap R) = N$ and must be that $5^2 \mid |N|$ so $|N| \in \{ 3\cdot 5^2, 7 \cdot 5^2, |G|\}$
if it's order is $3 \cdot 5^2$ or $7 \cdot 5^2$ by Cauchy there exists subgroups of order $3,7$ but we ruled those out thus $N = G$
contradiction