Why does the approximation for exponents $(a+b)^c \approx a^{c-bc} (a+1)^{cb}$ work?

Write your approximation $$a^{c-bc} \cdot (a+1)^{cb}=a^c\left(1+\frac 1a\right)^{bc}$$ and your approximation is $$\left(1+\frac 1a\right)^{b}\approx 1+\frac ba$$ Which is the first two terms of the binomial expansion. It will be reasonably accurate when $\frac ba \ll 1$ The next term is $\frac {b(b-1)}{2a^2}$

As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,

$$(a+b)^c ≈ a^{c-bc} \cdot (a+1)^{cb}$$

becomes

$$c \log(a+b) \approx c(1-b) \log(a) + cb \log(a+1)$$

or just

$$\log(a+b) \approx (1-b) \log(a) + b \log(a+1).$$

This is equivalent to doing a linear interpolation of $\log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $\log x$ will be close to linear between $a$ and $a+1$.