Look mum, we went from 108 to π in less than 23 iterations ;)

108 = 2^2 x 3^3 or the solution for ((x-1)^(x-1))(x^x) when x = 3

From which, and with no great leap of imagination at all, we can get the expression (3^(1÷3))^−(2^(1÷2)) or [ ((x^(1/x))^-((x-1)^(1/(x-1)) ] which is almost the lower bound of the expansion so with a bit of iteration we find that the lower bound occurs when x = π and n = 2.

(12^(1÷12))^−(11^(1÷11))^−(10^(1÷10))^−(9^(1÷9))^−(8^(1÷8))^−(7^(1÷7))^−(6^(1÷6))^−(5^(1÷5))^−(4^(1÷4))^−(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1)) = 0.841236826
(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1) = 0.771865166

(3^(1÷3))^−(2^(1÷2)) = 0.595775834

(4^(1÷4))^−(3^(1÷3)) = 0.606624308

(5^(1÷5))^−(4^(1÷4)) = 0.63430981

(3.13^(1÷3.13))^−(2.13^(1÷2.13)) = 0.594577953

(3.12^(1÷3.12))^−(2.12^(1÷2.12)) = 0.59462546

(3.11^(1÷3.11))^−(2.11^(1÷2.11)) = 0.594679901

(3.14^(1÷3.14))^−(2.14^(1÷2.14)) = 0.59453723

(π^(1÷π))^−((π−1)^(1÷(π−1))) = 0.594531362

A little bird helped me along the way x

So I guess I'd better figure out what the upper bound is now, though I assume it's =1

I do hope the upper bound is less than 1.

There's something really real about the interval between 0.594531362 and 0.841236826 or slightly higher.

But, alas;

(9999^(1÷9999))^−(9998^(1÷9998)) = 0.99907846

and

(9999^(1÷9999))^−(9999^(1÷9999)) is 1.

If you'd like to know where the original expression goes, the answer is nowhere special with respect to π:

((π−1)^(π−1))(π^π) = 186.271338549

((π−2)^(π−2))((π−1)^(π−1)) = 5.942346492

and know that it tends towards infinity and sub zero (but not less than -1)..