What is the 'scale factor' equation for a dark-matter dominated universe? The Friedmann equations can be solved exactly in presence of a perfect fluid with equation of state
$${\displaystyle p=w\rho c^{2}} \qquad p=w\rho c^2$$
where ${\displaystyle p}$ is the pressure, ${\displaystyle \rho }$  is the mass density of the fluid in the co-moving frame and $w$ is some constant.
In spatially flat case ($k = 0$), the solution for the scale factor is
$${\displaystyle a(t)=a_{0}\,t^{\frac {2}{3(w+1)}}}$$
where ${\displaystyle a_{0}}$ is some integration constant to be fixed by the choice of initial conditions. This family of solutions labelled by ${\displaystyle w}$ is extremely important for cosmology. E.g. ${\displaystyle w=0}$ describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as
$${\displaystyle a(t)\propto t^{2/3}}$$ matter-dominated
Another important example is the case of a radiation-dominated universe, i.e., when ${\displaystyle w=1/3}$. This leads to
$${\displaystyle a(t)\propto t^{1/2}}$$ radiation dominated
Note that this solution is not valid for domination of the cosmological constant, which corresponds to an ${\displaystyle w=-1}$. In this case the energy density is constant and the scale factor grows exponentially.
So, '$a$' is proportional to $t^{2/3}$ or $t^{1/2}$ for matter- or radiation-dominated universes, respectively... But if '$w$' is negative-one then '$a$' is proportional to $t^t$?
I mean, what is the exponent in this 'exponential growth' phase where the '$w$' 'constant' is $-1$?
You answered your question actually. You have said

Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1

then you are saying

But if 'w' is negative-one then 'a' is proportional to tt

The $w=-1$ does not apply to $a(t) \propto t^{2/3(1+w)}$