Show that $n(t+1)=(1+r)e^{-\alpha n(t)}n(t)$ is equivalent to $n(t+1)=(1+r)^{1-\frac{n(t)}{K}} n(t)$ where $K=\frac{\ln (1+r)}{\alpha}$

Question

Show that $n(t+1)=(1+r)e^{-\alpha n(t)}n(t)$ is equivalent to $n(t+1)=(1+r)^{1-\frac{n(t)}{K}} n(t)$ where $K=\frac{\ln (1+r)}{\alpha}$

I also have to show that $n(t+1)$ will grow when $r<0$ and the initial population is greater than $n=k$. I have also tried to show this but using both variations of the equation I get an increasingly negative value.