A formula for the power sums: $1^n+2^n+\dotsc +k^n=\,$? Is there explicit formula for the expression $1^n + 2^n + \dotsc + k^n\,$?
I know that for $n=1$ the explicit formula becomes $S=k(k+1)/2$ and for $n=3$ the formula becomes $S^2$. But what about general $n$?
I know there is a way using the Taylor expansion of $f(x)=1/(1-x)=1+x+x^2+\dotsc\;$, by differentiating it and then multiplying by $x$ and then differentiating again. Repeating this $n$ times, we get
$$\frac{d}{dx}(x\frac{d}{dx}(\dots x\frac{d}{dx}f(x))\dots )=1+2^nx^n+3^nx^n\dots.$$
Now do the same process but with the function $g(x)=x^{k+1}f(x)$. Then subtract them and we get $1+2^nx^n+\dots k^nx^n$. Because we have the explicit formulas $f(x)$ and $g(x)$ we can find the explicit formula by this process for arbitrary $n$. A big problem is that as $n$ grows, it is going take a lot of time finding the explicit formula. My question is therefore: are there other ways?

There is a result which does not depend on the Bernouli numbers or the Eulerian numbers. I found it in my first year at university, so you can be sure it involves nothing too complicated.
The method is that of discrete calculus, first I define $ \Delta f(x) = f(x+1)-f(x) $ and $ \Sigma f(x) = \sum\limits_{k=0}^{x-1} f(k) $ as discrete equivalents of differentiation and integration respectively.
The result will follow from a discrete equivalent of integration by parts, which I first proove:
$ f(x)g(x) = \Sigma f(x+1)g(x+1) -\Sigma f(x)g(x) $
$ \hspace{30 pt} = \Sigma f(x+1)g(x+1) -\Sigma f(x)g(x) + \Sigma f(x)g(x+1) - \Sigma f(x)g(x+1) $
$ \hspace{30 pt} = \Sigma (f(x+1)-f(x))g(x+1) + \Sigma f(x)(g(x+1)-g(x)) $
$ \hspace{30 pt} = \Sigma (\Delta f(x))g(x+1) + \Sigma f(x) \Delta g(x) $
$ \therefore \space \Sigma f(x) \Delta g(x) = f(x)g(x) - \Sigma (\Delta f(x))g(x+1) $
The expression we want a formula for will be a polynomial one degree higher than the power n, so we choose f(x) and g(x) with consideration that $ \deg(f(x)g(x)) $ should be $ n+1 $.
For a dash of simplicity, we can get $ f(x) \Delta g(x) = x^{n} $ by choosing $ f(x) = x^{n-1} $ and $ g(x) = \frac{1}{2} x(x-1) $.
For the other terms in the equivalence we need:
$ f(x)g(x) = \frac{1}{2} (x-1) x^n $
$ (\Delta f(x))g(x+1) = ((x+1)^{n-1}-x^{n-1})(\frac{1}{2} x(x+1)) = \frac{1}{2} x(x+1) \sum\limits_{m = 0}^{n-2} \binom{n-1}{m} x^m $
$ \hspace{57 pt} = \frac{1}{2} x(x+1)((n-1) x^{n-2} + \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} x^m) $
$ \hspace{57 pt} = \frac{1}{2}(n-1) x^n + \frac{1}{2}(n-1) x^{n-1} + \frac{1}{2} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (x+1) x^{m+1} $
Finally, plugging these in and rearranging:
$ \sum\limits_{k=0}^{x-1} k^n = \frac{1}{2} (x-1) x^n - \sum\limits_{k=0}^{x-1} \left( \frac{1}{2}(n-1) k^n + \frac{1}{2}(n-1) k^{n-1} + \frac{1}{2} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} \right) $
$ 2\sum\limits_{k=0}^{x-1} k^n = (x-1) x^n - (n-1) \sum\limits_{k=0}^{x-1} k^n - (n-1) \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} $
$ (n+1) \sum\limits_{k=0}^{x-1} k^n = (x-1) x^n + (1-n) \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} $
$ \sum\limits_{k=0}^{x-1} k^n = \frac{x-1}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
$ \therefore \sum\limits_{k=1}^{x} k^n = x^n + \sum\limits_{k=0}^{x-1} k^n = x^n + \frac{x-1}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
$ \hspace{31 pt} = \frac{x+n}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=1}^{x-1} k^{n-1} - \sum\limits_{k=1}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
Note that the power of k in the sums on the RHS does not exceed $ n-1 $.
Example with $ n = 3 $; the formula simplifies down to: 
$$ \sum\limits_{k=1}^{x} k^3 = \frac{1}{4} (x^4 + 3x^3 -3 \sum\limits_{k=1}^{x-1} k^2 - \sum\limits_{k=1}^{x-1} k) $$
Which further simplifies to the correct polynomial. Only the partial sums for $k^2$ and $k$ need be known, and this formula will provide that of any higher degree.
