How to simplify $$2^{a\left(2^{bc}-1\right)}-1\mod{N}$$
Where a,b,c are big integers with relatively small primes and N is a big integer that is hard to factor(it does not have small primes).
I have been trying doing the following:
$$2^{a\left(2^{bc}-1\right)}-1\mod{N}=$$ $$2^{a^{\left(2^{bc}-1\right)}}-1\mod{N}=$$ $$2^{\frac{a^{2^{bc}}}{a}}-1\mod{N}=$$ $$4^{\frac{a^{bc}}{a}}-1\mod{N}=$$
I don't know how to continue from here. Is there somehow a way to apply mod $N$ on just one part of $4^{\frac{a^{bc}}{a}}-1\mod{N}=$