I was able to do the first part of the question, in the second part (Proof by Induction),
I showed it holds for $n=1$:
Then I Assumed its true for $n=k$.
$$\begin{align}I_{k+1}=\frac{e^2}{2}-\frac{(k+1)}{2}I_k\end{align}$$
Then for $n=k+1$:
$$\begin{align}I_{k+2}=\frac{e^2}{2}-\frac{(k+1)+1}{2}I_{k+1}=\frac{e^2}{2}-(\frac{(k+1)+1}{2})(A_{k+1} e^2+B_{k+1})\end{align}$$
I don't know how to proceed.
You wrote you showed it holds for $n=1$.
Assume that $I_k=A_ke^2+B_k$ where $A_k,B_k$ are rational numbers.
Then, $$\begin{align}I_{k+1}&=\frac 12e^2-\frac 12(k+1)I_k\\&=\frac 12e^2-\frac 12(k+1)(A_ke^2+B_k)\\&=\left(\frac{1}{2}-\frac{(k+1)A_k}{2}\right)e^2+\left(-\frac{(k+1)B_k}{2}\right).\end{align}$$ Since $$\frac{1}{2}-\frac{(k+1)A_k}{2},\ \ -\frac{(k+1)B_k}{2}$$ are rational number, $I_{k+1}$ is of the form $A_{k+1}e^2+B_{k+1}$ where $A_{k+1},B_{k+1}$ are rational numbers.
