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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}.
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}?