Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

How Many Solutions?

Find all the solutions to the this equation.

Growing

Stage: 5 Challenge Level:

Prove that $(1 + \frac{1}{n})^n \leq e < 3$.

Which is larger:

(a) $1.000001^{1000000}$ or $2$?

(b) $100^{300}$ or $300!$ (i.e. 300 factorial)?