Today morning I had thought of a wonderful way to calculate the general sum of $$ \sum_{i=0}^{i=n} i^n$$
Using just things taught in elementary high school calculus. So, my method is as follows,
First consider the geometric series:
$$ S = \sum_{i=0}^{i=n} x^i$$
from the formula for geometric series,
$$ \frac{x^{n+1} - 1}{x-1} = \sum_{i=0}^{i=n} x^i$$
Rewriting this into the form in which leibniz product rule is more applicable,
$$ \sum_{i=0}^{i=n} x^i = (x^{n+1} - 1) (x-1)^{-1}$$
take a derivative on both sides,
$$ \sum_{i=0}^{i=n} i x^{i-1} = \frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]$$
If I take limit as x goes to one on both sides I get formula for sum of first n integers,
now suppose I multiply both sides by x and take derivative,
$$ \sum_{i=0}^{i=n} i^2 x^{i-1} = \frac{d}{dx} [x\frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]]$$
Taking limit as $ x \to 1$ gives me formula for first n squares,
now do same procedure again,
$$ \sum_{i=0}^{i=n} i^3 x^{i-1} = \frac{d}{dx} \left[x\frac{d}{dx} \left[x\frac{d}{dx} [ (x^{n+1} - 1) (x-1)^{-1}]\right]\right]$$
To generalize, let,
$$ P = x \frac{d}{dx}$$
And,
$$ P \cdot P =P^2 = x \frac{d}{dx} [ x \frac{d}{dx} ]$$
Then,
$$ \sum_{i=0}^{i=n} i^k = \lim_{x \to 1} P^k [ (x^{n+1} - 1) (x-1)^{-1}]$$
Is there a way to generally evaluate the limit? Thank you.