How can we show that $n < 2^n$ for all natural numbers $n$?

Since you ask for proof other than that by induction, consider $$2^n=(1+1)^n=\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n} > \binom{n}{1}=n$$ as an application of Binomial theorem.

$2^n$ is the cardinality of the power set of a set of $n$ elements.

The power set includes $n$ singletons and the empty set.

Summary of comments that should have been answers:

• If $n<2^n$ with $n\ge1$ then $n+1\le 2n<2^{n+1}$, so we can induct from $n=1$ onward, and only need to check the base steps $n=0$ and $n=1$;
• A hypothetical function $f$ mapping elements of $\{1,\,\cdots,\,n\}$ to subsets thereof can't satisfy $f(k)=\{m|m\notin f(m)\}$ (as this would lead to the contradiction $k\in f(k)\iff k\notin f(k)$), so there are too many subsets to pair with elements.

A recent comment reveals the first approach is known & not desired.