Ironic Isn't It?

Let $a^b = c^d = e^f = g^h = i^j = k^l = M$ for some positive integer $M$ .

The value of $M$ must be in the form of $m^n$ , so $a = m^{n/b}, c = m^{n/d} , \ldots, k = m^{n/l}$

And because $a,c,e,g,i,k$ are distinct, $m \geq 2$ with $a$ greater than $c,e,g,i,k$ implies $n/b \geq 6 \Rightarrow \min (a) = 2^6 = 64$

By finding the lowest common multiple: $\text{lcm}(1,2,3,4,5) = 2^2 \times 3 \times 5 = 60$ , we have $d = 60, f = 30, h = 20, j = 15, k = 12$ . And $a^b = c^d \Rightarrow 64^b = 2^{60} \Rightarrow b = 10$ which satisfy the constraint above. Hence $a = \boxed{64}$

We hope to find $x > 6$ with $\text{lcm} (x,6,5,4,3,2,1) \geq 60$ , but trial and error shows this is the bare minimum, so the equation is indeed minimized.