Is $\pi$ more transcendent than $e$? I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?
This question at MathOverflow has several answers discussing several senses in which one can say that one real number is more irrational than another, including irrationality measures and other hierarchies of complexity for real numbers.
This might be a subset of the previous answers but I think this is still worth writing it as a separate answer.

One reason to believe $e$ is less transcendental than $\pi$ is that there is a nice pattern in the continued fraction of $e$ as opposed to that of $\pi$.

$e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,16,\ldots]$ whereas there is no such pattern for $\pi$. (We can find patterns in $\pi$ as well if we look at generalized continued fractions where the "numerators" in the continued fraction need not be $1$ always)