The least value of $b$ such that $2^{3^{\cdots^a}}\leq b^{(b-1)^{\cdots^2}}$ I'm interested in an upper bound for the minimal positive integer $b$ for which $$2^{3^{4^{\cdots^a}}}\leq b^{(b-1)^{\cdots^{3^{2}}}}$$ holds, given a positive integer $a\geq2$.
If possible, but I bet it's not, a closed formula for $b$ would be great...

A natural but rather unuseful first step would be to take logarithms some times:
$$3^{4^{\cdots^{a}}}\ln(2)\leq(b-1)^{\cdots^{3^2}}\ln(b)$$
But after doing this two more times one will obtain a sum in the logarithm and things get ugly.
Any estimation of $b$ would be interesting.

Edit:
Thank's to Ross Millikan's answer below, I'm somehow expecting the answer will be $b=a+1$, but until now this is not much more then intuition.

If you want an estimate, mine would be $b=a+1$  The most important thing about power towers is the number of levels-nothing can compete with that.  For the same number of levels ($a=b$), your stacking on the left is clearly greater, so $b \gt a$.  Your idea of taking logs is a good one.  Then if you recognize that with those towers around, the difference between $\log 2$ and $\log b$ is trivial, you can cancel them and take another log.  Keep going.  Clearly this is a huge handwave, but one I believe in.
