jmc

Note that $p(x)$ takes on the same values as $x^2+1$ for $x=1,$ 2, 3, and 4. So, let $$q(x)=p(x)-x^2-1.$$Then $q(x)$ is also a monic quartic polynomial. Also, $q(1)=q(2)=q(3)=q(4)=0,$ so $$q(x)=(x-1)(x-2)(x-3)(x-4).$$Hence, $p(x)=(x-1)(x-2)(x-3)(x-4)+x^2+1.$ We can set $x=5,$ to get $p(5)=\boxed{50}.$