Problem 5

Suppose $p>a$. Then $p\ge a+1$ and the function $a^t-t$ is increasing for integers $t\ge a+1$, so $b!=a^p-p\ge a^{a+1}-(a+1)>a!$. Hence $b>a$. Reducing $b!=a^p-p$ modulo $a$ gives $a\mid p$, impossible because $1<a<p$; the case $a=1$ does not satisfy the original equation. Thus $p\le a$.