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If 1*1!+2*2!+3*3!+ . . .+n*n ! =(n+1)!-1...

If `1*1!+2*2!+3*3!+ . . .+n*n ! =(n+1)!-1` then show that, `1*1!+2*2!+3*3!+ . . . +n*n! +(n+1)(n+1)! =(n+2)!-1`

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