Exponential Space Complexity equality
Suppose first that $\log n$ stands for base 2 logarithm. In this case, since $n = 2^{\log n}$, we have
$$
2^{c\log^2 n} = (2^{\log n})^{c\log n} = n^{c \log n},
$$
from which your identity immediately follows.

If $\log n$ uses some other base, we still get the same result, because both sides of the identity are insensitive to the base of the logarithm. This is because different logarithms differ by a constant factor, and this constant factor can be swallowed by the union over $c$. Details left to you.