Function under a function

Suppose $1 \geq f(1).$ Then since $f$ is increasing, $1 \geq f(1) \geq f(f(1)) = 1! + 2! = 3$ which is a contradiction, so we may conclude $1<f(1).$

Since again $f$ is increasing, $1 < f(1) < f(f(1)) = 3$ which implies $f(1)=2$ and $f(2) =3.$ Then $f(26) = f(2! + 4!) = f(f(f(2))) = f(f(3)) = 3! + 6! = \boxed{726}$