Hint: We can write $$2^{n-2}\cdot n<4\cdot n!$$ by cross multiplication and this is $$2^{n-2}<(n-1)!$$, which is easier to prove. and we have to prove that $$2^{n-1}<n!$$ if $$2^{n-2}<(n-1)!$$ multiplying this inequality by $2$ we get
$$2^{n-1}<2(n-1)!$$ and $$2(n-1)!<n!$$ for $$n>2$$
Hint: We can write $$2^{n-2}\cdot n<4\cdot n!$$ by cross multiplication and this is $$2^{n-2}<(n-1)!$$, which is easier to prove. and we have to prove that $$2^{n-1}<n!$$ if $$2^{n-2}<(n-1)!$$ multiplying this inequality by $2$ we get $$2^{n-1}<2(n-1)!$$ and $$2(n-1)!<n!$$ for $$n>2$$