1. Let's state the problem clearly: If a set $S$ has $n$ elements, we want to prove that its power set $\mathcal{P}(S)$ has $2^n$ elements. 2. Recall that the power set of $S$ is the set of all subsets of $S$, including the empty set and $S$ itself. 3. We prove this by induction on $n$. 4. **Base case:** When $n=0$, $S$ is the empty set. The power set of the empty set is $\{\emptyset\}$, which has exactly 1 element. 5. Note that $2^0=1$. So the statement holds for $n=0$. 6. **Inductive hypothesis:** Assume that for a set with $k$ elements, the power set has $2^k$ elements. 7. **Inductive step:** Consider a set $S$ with $k+1$ elements. Take any element $a \in S$ and let $T = S \setminus \{a\}$, which has $k$ elements. 8. By the inductive hypothesis, $\mathcal{P}(T)$ has $2^k$ elements. 9. Every subset of $S$ either contains $a$ or does not contain $a$. 10. The subsets of $S$ that do not contain $a$ are exactly the subsets of $T$, so there are $2^k$ such subsets. 11. The subsets of $S$ that do contain $a$ can be formed by taking each subset of $T$ and adding $a$, so there are also $2^k$ such subsets. 12. Therefore, total subsets in $\mathcal{P}(S)$ equals $2^k + 2^k = 2 \times 2^k = 2^{k+1}$. 13. This completes the inductive step; hence by induction, a set with $n$ elements has a power set with $2^n$ elements. **Final answer:** The power set of a set with $n$ elements has $2^n$ elements.