sum_i=1..n (2i-1) = n²

Out of curiosity, I tried to find an analogous decomposition for n³. After some poking around, I found this:

sum_i=1..n (3i²-3i+1) = n³

I noticed that the coefficients of the summed term seem suspiciously familiar:

2, 1

3,-3, 1

This looks like the 2nd and 3rd rows of Pascal's triangle, except without the initial entry, and in alternating signs. To test my hypothesis, I asked WolframAlpha to solve the following sum:

sum_i=1..n (4i³-6i²+4i-1)

Guess what the answer was? That's right: n⁴.

The next sum continues this pattern:

sum_i=1..n (5i⁴-10i³+10i²-5i+1) = n⁵

Looks like my hypothesis was right: n

^{e}can be written as a sum of lower powers of n in alternating signs, the coefficients of which are taken from the n'th row of Pascal's Triangle except without the first entry. (The first row also trivially holds: sum_i=1..n (1) = n¹.)

Why this is so, however, eludes me. On the surface it looks suspiciously like solving for n

^{e}in the expansion of (n-1)

^{e}, however, that doesn't account for the summation sign. The formula without the summation, obviously, does not hold. So why is it that n'th powers can be decomposed exactly in this way? Anybody can explain why this is so?