$\text{A Biquadratic}^{\text{polynomial}}$

From the given equations, $f(x) - 1 = p(x)(x - 1) = q(x)(x - 2) = r(x)(x - 3) = s(x)(x - 4)$ , so four factors of $f(x) - 1$ must include $(x - 1)$ , $(x - 2)$ , $(x - 3)$ , and $(x - 4)$ .

If $f(x)$ is a monic biquadratic polynomial, then so is $f(x) - 1$ , which means $f(x) - 1$ can only have up to four factors, the ones given above.

Therefore, $f(x) - 1 = (x - 1)(x - 2)(x - 3)(x - 4)$ , which means $f(x) = (x - 1)(x - 2)(x - 3)(x - 4) + 1$ , and $f(5) = (5 - 1)(5 - 2)(5 - 3)(5 -4) + 1 = \boxed{25}$ .

$\Large{Analyzing\space only\space one\space polynomial}$

$p(x):$ $f(2)=p(2)(2-1)+1=q(2)(2-2)+1$ $\Rightarrow p(2)\cancel{+1}=\cancel{1}$ $\Rightarrow \boxed{p(2)=0}..........[A_1]$ $f(3)=p(3)(3-1)+1=r(3)(3-3)+1$ $\Rightarrow 2p(3)\cancel{+1}=\cancel{1}$ $\Rightarrow \boxed{p(3)=0}..........[B_1]$ $f(4)=p(4)(4-1)+1=s(4)(4-4)+1$ $\Rightarrow p(4)\cancel{+1}=\cancel{1}$ $\Rightarrow \boxed{p(4)=0}..........[C_1]$ From $[A_1],[B_1]$ and $[C_1]$ and assuming that $p(x)$ is a monic polynomial we get $p(x)=(x-2)(x-3)(x-4)$ $\Rightarrow f(5)=p(5)(5-1)+1=(5-2)(5-3)(5-4)(4)+1=(3)(2)(1)(4)+1=25$

A request to @Mahdi Raza to mention that $p(x),q(x),r(x),s(x)$ are monic polynomials.

$f(x)=(x-1)(x-2)(x-3)(x-4)+1$

$\implies f(5)=(4)(3)(2)(1)+1=\boxed{25}$