Using mathematical induction, to prove that
`1*1!+2*2!+3.3!+ . . . .+n*n! =(n+1)!-1`, for all `n in N`

Explanation

Let P(n): 1.1!+2.2!+3.3!+….+n.n!=(n+1)-1
When n=1
LHS=1.1!=1
and RHS=(n+1)!-1
=2!-1
=2-1=1
`therefore LHS=RHS`. Hence P(1) is true …..(a)
Now, Let P(k) be true.
`Rightarrow 1.1!+2.2!+3.3!+....+k.k! =(k+1)!-1....(i) `
To prove P(k+1) is true.
i.e. `1.1!+2.2!+3.3!+....+(k+1)(k+1) =(k+2)!-1...(ii)`
Adding (k+1) (k+1)! to both sides of (1), we get
[1.1!+2.2!+3.3!+......+k.k!] +(k+1) (k+1)!
`(k+1)!+(k+1)(k+1)!-1`
=(k+1)!(k+1+1)-1
=(k+1)!(k+1)1-1
=(k+2)!-1 `[therefore (n+1)k!=(k+1)! "for all "n in N]`
Hence, P(k+1) is true, whenever P(k) is true. (b)
From (a) and (b) by the principle of mathematical induction it follows that P(n) is true for all natural number n.