Yes, there is!
You can refer to Hall and Knight for the formula.
If you don't have the book then write back and i'll post the formula.
With regards
Assume Sx=1x + 2x + 3x ...... + nx .
Then Sx-1=1x-1 + 2x-1 + 3x-1 ...... + nx-1
& S0= 10 + 20 + 30 ...... + n0 =1+1+1......+1=n
Let:
Z = [(n+1)x+1 -1] - [x+1C2 (Sx-1)+ x+1C3 (Sx-2)+ x+1C4 (Sx-3)...+ x+1Cx+1 (S0)]
Then Sx = Z /( x+1)
The general formula is difficult to use. U may find Sx progressively then substitute in Z where required.
Why do you need a book for that , all you have to do is just subtract the nth power from its n+1th power.Further you could use telescoping for the summation.But in general no one does ask you to these kind of things,all they ask is the proof giving you the formula. which is an obvious implication of mathematical induction. but anyways I shall give you my ways of doing things which you may find difficult to use but here goes
n^x-n-1^x=use the binomial exansion
n-1^x-n-2^x=Do the same
Continue this till you get 2^x-1^x & sum the left hand terms & the right hand terms you should get the answer but believe me you dont want to try this out. you see this way can only be opted for lower x.but if your x is big forget it!