We want to find equivalence of the expression
$$e^{n\log(n)-(n+e)\log(n + e)}$$
Note that: $$\log(n+t)=\log\left[n\left(\frac{t}{n}+1\right)\right]=\log(n) + \frac{t}{n} +o\left(\frac{t}{n}\right)$$ Can we say that $$n\log(n+t)=n\log(n) + t +o(t),$$ and then $$u_n=e^{n\log(n)-(n+t)\log(n+t)}=e^{n\log(n)-n\log(n+t)-t\log(n+t)}\sim ?$$
The expression is the same as $$e^{n\log(n)-(n+e)\log(n+e)}=\frac{n^n}{(n+e)^{n+e}}=\left[\left(1+\frac{-e}{n+e}\right)^{\frac{n+e}{-e}}\right]^{\frac{-e}{n+e}n}\frac{1}{(n+e)^e}\sim \frac{e^{-e}}{n^e}$$
Is that a simple enough expression?