We have $d/dx(b^x) = b^x\log(b)$, with $b>0$.
If $b < e$, then $b^x > b^x\log(b)$. If $b = e$, then $b^x = b^x\log(b)$. If $b > e$, then $b^x < b^x\log(b)$. In fact, if you take the $Nth$ derivative, as $N$ approaches infinity, the limits are $0$, $b^x$, and $\infty$ respectively.
One might also notice that if you consider a function $a_{n+1} = d/dx(a_n)$, with $a_0=b^x$, then $a_n = b^x\log(b)^n$. This means $a_n$ grows exponentially for $b > e$, but decays for $b < e$.
The math is clear, but I'm having trouble finding the intuition as to why this is. In other words, what is so different about $2.7^x$, $e^x$, and $2.72^x$ that would cause such a drastic change in the acceleration of the derivatives?