Why does the base-emitter voltage of a BJT decrease with temperature?

According to Sedra/Smith Microelectronic Circuits, $v_{BE}$ changes by $-2\text{mV}/\text{°C}$. I cannot understand how this could possibly be the case given the equations I am familiar with.

With all currents kept constant, we have:

$\large{i_E = \frac{I_s}{\alpha}e^{v_{BE}/V_T}}$

To keep $i_E$ constant, any change in $V_T$ would have to be accompanied by a change of the same factor in $v_{BE}$, otherwise $\alpha$ or $I_s$ would have to change, which as far as I understand is not possible.

So, how can $v_{BE}$ be inversely proportional to $V_T$?

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$I_S$ is highly temperature dependent. As the temperature of the material increases, more electron-hole pairs are thermally generated, increasing $I_S$. Here's a link that gives the formula SPICE uses for $I_S$
$I_S(T_1)=I_S(T_0)\left[\dfrac{T_1}{T_0}\right]^{XTI}exp\left[{\dfrac{-E_gq(T_1T_0)}{k(T_1-T_0)}}\right]$
I believe that $I_S$ is roughly cubic in $T$