Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=0$, we have $LHS=c_0=3$ and $RHS=3^{2^0}=3$, thus $LHS=RHS$ and $P(0)$ is true.
3. Assume $P(k), k\gt0$, is true, that is $c_k=3^{2^k}$
4. For $n=k+1$, we have $LHS=c_{k+1}=(c_k)^2=(3^{2^k})^2=3^{2\cdot2^k}=3^{2^{k+1}}=RHS$
5. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.