Note that your informal argument is wrong too. $(2n+1)$ is greater than $(n+1)$, but
$ \lim_{n \to +\infty} \frac{n+1}{2n+1} = \frac{1}{2} \neq 0 $
What you mean was not that $2^{n \cdot n!}$ is merely greater, but some superlative statement about how much greater it is.
Formally, you're trying to say that $2^{n \cdot n!} \in \omega(n+1)$, or conversely, $n+1 \in o(2^{n \cdot n!})$. These facts immediately imply the limit you seek. See wikipedia for what this means.
If you're familiar with asymptotics, then starting from the basic fact $n \in o(2^n)$, each step in the chain
$ n+1 \in o(2^{n+1}) = o(2^n) \subseteq o(2^{n \cdot n!}) $
is an easy argument to make.
If you're not familiar with asymptotics, you'll have to reproduce the theory to some extent. e.g. an inductive argument that $(2n+1) / 2^{2n+1} < 1/2^{n+1}$, by noting the numerator doesn't double when you increment $n$.