Question on Factorial Induction

1/2! + 2/3! + 3/4! + 4/5! + ... + n/(n+1)! = (n+1)!-1/(n+1)!
How can I prove this and second problem is to find the sum of
1/2! + 2/3! + 3/4! +4/5! + ... + 9/10! in rational form

Careful with the parentheses:
1/2! + 2/3! + 3/4! + 4/5! + ... + n/(n+1)! = ((n+1)! - 1)/(n+1)! = 1 - 1/(n+1)!.

Proof by induction:

Base Case.
This is true for n = 1, because 1/2! = 1/2 = 1 - 1/2!

Inductive Step.
Assuming that the claim is true for n = k,
1/2! + 2/3! + ... + k/(k+1)! + (k+1)/(k+2)!
= (1/2! + 2/3! + ... + k/(k+1)!) + (k+1)/(k+2)!
= [1 - 1/(k+1)!] + (k+1)/(k+2)!, via inductive hypothesis
= 1 - (k+2)/(k+2)! + (k+1)/(k+2)!
= 1 - 1/(k+2)!, thus completing the inductive step (the claim is true for n=k+1).
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Letting n = 9 yields
1/2! + 2/3! + 3/4! + 4/5! + ... + 9/10! = (10! - 1)/10! = 3628799/3628800.

I hope this helps!