I had some trouble with the first part:

Prove that e - (1+1/1!+1/2!+...+1/n!) < 1/(n!n) for all n in {1,2,...}

It was pretty easy to prove that the residual is less than (e-1)/(n+1)!, but that's not good enough. After a little trial and error, I found the following manipulations. (Of course I discovered the transformation starting at the bottom and working my way to the top.)

{n/(2n+6)}{1 + 1/(n+4) + (n+3)!/(n+5)! + (n+3)!/(n+6)! + ...} < (1/2)(e-1) < 1

The first inequality follows because

n/(2n+6) < 1/2

and

1 + 1/(n+4) + (n+3)!/(n+5)! + (n+3)!/(n+6)! + ... < 1 + 1/2 + 1/3! + 1/4! + ... = e-1

So, if we are prepared to accept the fact that (e-1)/2 < 1, then we can proceed...

{1/(n+3)}{1 + 1/(n+4) + (n+3)!/(n+4)! + ...} < 2/n

1/(n+3) + (n+2)!/(n+3)! + (n+2)!/(n+4)! + ... < 2/n

1 + 1/(n+3) + (n+2)!/(n+3)! + (n+2)!/(n+4)! + ... < 1 + 2/n = (n+2)/n

1/(n+2) + (n+1)!/(n+3)! + (n+1)!/(n+4)! + ... < 1/n

1 + 1/(n+2) + (n+1)!/(n+3)! + (n+1)!/(n+4)! + ... < (n+1)/n

{n/(n+1)}{1 + 1/(n+2) + (n+1)!/(n+3)! + ...} < 1

Therefore,

e - (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)
= (1/(n+1)! + 1/(n+2)! + ...)
= (1/n!n){n/(n+1)}{1 + 1/(n+2) + (n+1)!/(n+3)! + ...}
< (1/n!n)

Q.E.D.