I know there is a nice equation for $n!$, but is there one for $n^n$? I was thinking you could get it with the fact $n^n=a^{n\log_an}$ but I can't seem to make the needed jump.
Edit: It was suggested by goblin to use the fact that $(n+1)^{n+1} = (n+1)(n+1)^n$ and expand with the binomial theorem. This gives $a_{n+1} = (n+1)\sum_{i=0}^n \binom n {n-i} n^i$
Which is still a little more complicated then I'd like. I'd love to have something of the form $a_{n+1}=a_nb^{f(n,b)}$ for every $b$
Using the $W$ Lambert function (the inverse of $x\mapsto xe^x$) we get:
$$a_n=n^n$$ $$\ln a_n=n\ln n=\ln n e^{\ln n}$$ $$W(\ln a_n)=\ln n$$ Therefore $$a_{n+1}=[e^{W(\ln a_n)}+1]^{e^{W(\ln a_n)}+1}$$
On the other hand, from your equation $$(n+1)^{n+1}=n^n\,b^{f(b,n)}$$ we get $$f(b,n)=\log_b\frac{(n+1)^{n+1}}{n^n}=\log_b(n+1)+n\log_b\left(1+\frac{1}{n}\right)$$