Considering the sequence: $$b_1=1 \quad\text{and}\quad b_{n+1}=\left(1+\frac{1}{n}\right)^n\times b_n$$
Show by induction that $b_n=\frac{n^n}{n!}$.
So, what I've done so far is:
$$\begin{align} b_{n+1} &=\left(1+\frac{1}{n}\right)^n\times \frac{n^n}{n!} \\ &=\frac{\left(1+\frac{1}{n}\right)^n \times n^n}{n!} \\ &=\frac{\left(\frac{n+1}{n}\right)^n\times n^n}{n!} \\ &=\frac{\frac{(n+1)^n}{n^n}\times n^n}{n!} \\ &=\frac{(n+1)^n}{n!} \end{align}$$
I thought this was correct, but in the solution given with the exercise, it is said that $$b_{n+1}=\frac{(n+1)^{n+1}}{(n+1)\times n!}$$ I don't see why do I have to add this last part. If someone could help me, that would be great.
Thank you so much for your time. If something's not very clear, please let me know.
What you want is $b_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^{n}}{n!}\times\frac{n+1}{n+1}$. So your answer is right but you have to multiply by $n+1$ above and below the fraction.