Recursion Relation

\hspace{-16}\mathbf{\left(1+\frac{1}{a_{1}}\right).\left(1+\frac{1}{a_{2}}\right).\left(1+\frac{1}{a_{3}}\right)........\left(1+\frac{1}{a_{n}}\right)=\prod_{r=1}^{n}\left(1+\frac{1}{a_{r}}\right)}$\\\\\\ $\mathbf{=\prod_{r=1}^{n}\left(\frac{a_{r}+1}{a_{r}}\right)=\prod_{r=1}^{n}\left(\frac{a_{r+1}}{a_{r}}\right).\left(\frac{1}{r+1}\right)}$\\\\\\ Using Recursive relation $\mathbf{a_{n}=n.(1+a_{n-1})}$\\\\ Put $\mathbf{n=r+1},$ We Get $\mathbf{a_{r+1}=(r+1).(1+a_{r})}$\\\\ $\mathbf{(1+a_{r})=\frac{a_{r+1}}{r+1}}$\\\\\\ So $\mathbf{\prod_{r=1}^{n}\left(\frac{a_{r+1}}{a_{r}}\right).\left(\frac{1}{r+1}\right)=\frac{a_{n+1}}{(n+1)!}}$ , bcz $\mathbf{a_{1}=1}$ (Given)\\\\\\ Now Let $\mathbf{b_{n+1}=\frac{a_{n+1}}{(n+1)!}}$\\\\\\ Then $\mathbf{b_{n}=\frac{a_{n}}{n!}}$\\\\\\ So $\mathbf{b_{n+1}-b_{n}=\frac{a_{n+1}}{(n+1)!}-\frac{a_{n}}{(n)!}=\frac{(n+1).(1+a_{n})}{(n+1)!}-\frac{a_{n}}{(n)!}=\frac{1}{(n)!}}$\\\\\\

\hspace{-16}$ So $\mathbf{b_{n+1}-b_{n}=\frac{1}{(n)!}}$\\\\\\ Put $\mathbf{n=1,2,3,4,5........}$ and Added..........$\\\\\\ \mathbf{\sum_{r=1}^{n}\left(b_{r+1}-b_{r}\right)=\sum_{r=1}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{b_{n+1}-b_{1}=\sum_{r=1}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{b_{n+1}=b_{1}+\sum_{r=1}^{n}\frac{1}{r!}=\sum_{r=0}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{\lim_{n\rightarrow \infty} b_{n+1}=\lim_{n\rightarrow \infty}\frac{a_{n+1}}{(n+1)!}=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\frac{1}{r!}=e}$