Answer
See below.
Work Step by Step
1. Let $P(n)$ be the statement to be proved.
2. For $n=1$, we have $LHS=d_1=2$ and $RHS=\frac{2}{1!}=2$, thus $LHS=RHS$ and $P(1)$ is true.
3. Assume $P(k), k\gt1$, is true, that is $d_k=\frac{2}{k!}$
4. For $n=k+1$, we have $LHS=d_{k+1}=\frac{d_k}{k+1}=\frac{\frac{2}{k!}}{k+1}=\frac{2}{(k+1)!}=RHS$
5. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.