For $T(n) = 16(T/4) + n^2\lg^3n$ prove: $T(n) = \Theta(n^2\lg^3n)$
What you want to prove is false.

$T(n) = 16 T(n/4) + n^2\log^3 n$ has solution $T(n) = \Theta(n^2 \log^4 n)$.

To see this notice that this recurrence fits in the general form $T(n) = a T(n/b) + f(n)$ once you set $a=16, b=4, f(n)=n^2\log^3 n$.

Since $f(n) = n^2 \log^3 n = \Theta( n^{ \log_b a } \cdot \log^k n)$ for $k=3$, you can apply the Master theorem to conclude that $T(n) = \Theta(n^{ \log_b a } \cdot \log^{k+1} n)$.