Answer
See explanations.
Work Step by Step
Step 1. Prove the statement is true when $n=1$: $1\lt2^1$ , thus it is true for $n=1$.
Step 2. Assume the statement is true when $n=k$: we have $k\lt 2^k$ or $k-2^k\lt 0$
Step 3. Prove it is true for $n=k+1$: $(k+1)-2^{k+1}=k+1-2^k-2^k=(k-2^k)+(1-2^k)$. As $k\geq1$, we have $2^k\gt1$ or $1-2^k\lt0$, we also have $k-2^k\lt 0$ and the sum of two negative numbers will also be negative, so that $(k+1)-2^{k+1}\lt0$ or $(k+1)\lt 2^{k+1}$
Thus, the statement is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n$.
