Baltic Way shortlist

Since $n!$ is divisible by $3^{\lfloor n/3\rfloor }$, $(2^{a-b}-1)2^b=2^a-2^b$ is divisible by $3^{\lfloor n/3\rfloor }$. From the Lifting the Exponent Lemma we obtain that $a-b$ is divisible by $3^{\lfloor n/3\rfloor -1}$. So $a-b\ge 3^{\lfloor n/3\rfloor -1}\ge 3^{n/3-2}$. Hence, the right hand side of our prior equality is greater than $$2^{a-b}-1\ge 2^{a-b-1}\ge 2^{3^{n/3-2}-1}$$ which for sufficiently large $n$ is obviously greater than $2^{n^2}>n!$.