I am afraid that no. Since for fixed $k$ we have $(n+k)^{n+k}\sim n^n\cdot n^ke^k$, when we divide the recurrence by $n^n$, we get several terms equivalent to polynomials in $n$,and nothing cancels, since $e$ is not algebraic.

I am afraid that no. Since for fixed $k$ we have $(n+k)^{n+k}\sim n^n\cdot n^ke^k$, when we divide the recurrence by $n^n$, we get several terms equivalent to polynomials in $n$ (namely, to $e^kn^kP_k(n)$ for $k=0,1,\ldots,d$), and the leading terms of these polynomials do not cancel, since $e$ is not algebraic.

I am afraid that no. Since for fixed $k$ we have $(n+k)^{n+k}\sim n^n\cdot n^ke^k$, when we divide the recurrence by $n^d$ we get several terms equivalent to polynomials in $n$ which do not cancel out since $e$ is not algebraic.

I am afraid that no. Since for fixed $k$ we have $(n+k)^{n+k}\sim n^n\cdot n^ke^k$, when we divide the recurrence by $n^n$, we get several terms equivalent to polynomials in $n$,and nothing cancels, since $e$ is not algebraic.