The cases for $n=1$ and $n=2$ follow by inspection.

For even $n>2$ the case follows immediately from the case for $n-1$ since $n$ is not prime.

So let $n=2m+1$ with $m>0$ and consider $(1+1)^{2m+1}$ and its binomial expansion. Since $\displaystyle {{2m+1}\choose m}={{2m+1}\choose{m+1}}$ and each term occurs exactly once, it follows that $\displaystyle {{2m+1}\choose m}\leq 4^m$ Each prime $p$ with $m+1<p\leq 2m+1$ divides ${2m+1}\choose m$ implying that their product also divides $\displaystyle {{2m+1}\choose m}$ Hence $$ \vartheta(2m+1)-\vartheta(m+1)\le\log{{2m+1}\choose m}\le m\log 4. $$ By the induction hypothesis, $\vartheta(m+1)\leq(m+1)\log4$ and so $\vartheta(2m+1)\leq(2m+1)\log4$