Answer: \(2!-S_1=1,\quad 3!-S_2=1,\quad 4!-S_3=1,\quad 5!-S_4=1.\)

A suitable conjecture is \(S_n=(n+1)!-1\).

Hence, for all positive integers \(n\), \(S_n=(n+1)!-1\).

Since \(u_r=r\times r!\),

\(S_1=1\times 1!=1\)

\(S_2=1\times 1!+2\times 2!=1+4=5\)

\(S_3=1+4+3\times 3!=1+4+18=23\)

\(S_4=23+4\times 4!=23+96=119\)

So

\(2!-S_1=2-1=1\)

\(3!-S_2=6-5=1\)

\(4!-S_3=24-23=1\)

\(5!-S_4=120-119=1\)

This suggests the conjecture

\(S_n=(n+1)!-1\).

We now prove this by induction.

Let \(H_n\) be the statement \(S_n=(n+1)!-1\).

Base case: \(n=1\).

\(S_1=1\), and \((1+1)!-1=2!-1=2-1=1\).

So \(H_1\) is true.

Inductive step: Assume \(H_k\) is true for some positive integer \(k\), so

\(S_k=(k+1)!-1\).

Then

\(S_{k+1}=S_k+u_{k+1}\)

\(=(k+1)!-1+(k+1)(k+1)!\)

\(=(k+1)![1+(k+1)]-1\)

\(=(k+1)!(k+2)-1\)

\(=(k+2)!-1\).

So \(H_{k+1}\) is true.

Therefore, by mathematical induction,

\(S_n=(n+1)!-1\) for all positive integers \(n\).