Is there a Definite Integral Representation for $n^n$? The factorial $n!$ has a nice representation as definite integral:
$$
    n!=\Gamma(n+1)=\int_0^\infty t^{n} e^{-t}\, \mathrm{d}t. \!  
$$
Is it possible to write down such an integral for $n^n$ as well?
I tried to come up with an integral that reproduces a $n$ factor, $n$-times, but without success. I don't see a way to stop the partial integration process like in the $n!$ case. So this might not work here and I currently can't think of another way. If it helps to restrict $n$, feel free to do so.
The only thing a found online so far is the Lambert's $W$ function, which is involved when solving $x^x=z$, but I'm not sure if this helps. 
EDIT: Answers with integrals of the form $\displaystyle n^n=\int_0^\infty \cdots dt$ are preferred.
 A: This may not be the line of thinking you are after, but it does give the integral you desire.
$$n^n=\int_0^n nx^{n-1}\ dx$$
A: $$\frac{\Gamma (\alpha)}{s^{\alpha}}=\int_{0}^{\infty }t^{\alpha-1}e^{-st}dt\tag{1}$$
$$\frac{\Gamma (s)}{s^{s}}=\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{2}$$
$$s^{-s}=\frac{1}{\Gamma (s)}\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{3}$$
$$s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}\tag{4}$$
A: No need for the Lambert W function , respecting your limits and using $\Gamma$ we get that
$$ n^{n}=\frac{1}{\Gamma (n+1)}\int_{0}^{\infty }e^{-\frac{t^{ \frac{1}{n}}}{n}} dt $$
Update:
Lookup the Exponential Integral and its relationship with the Incomplete gamma function
$$E_{n}(x)=x^{n-1}\Gamma(1-n,x)\tag{1}$$
for $n=1-n$ and $x=\frac{1}{n}$ we get that :
$$E_{1-n}(\frac{1}{n})=\int_{1}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt =  n^{n}\Gamma (n,\frac{1}{n})\tag{2}$$
Changing the integration limits to $[0,1]$ : 
$$\int_{0}^{1}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt  =  n^{n}(\Gamma(n)-\Gamma (n,\frac{1}{n}))\tag{3}$$ 
Combining 2+3 we get the solution:
$$n^{n}=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt$$
The top result is due to the relationship :
$$n E_{1-n}(\frac{1}{n})= \int_{1}^{\infty} e^{-\frac{t^{\frac{1}{n}}}{n}} dt\tag{4}$$
A: The correct question (which someone has just asked me) may be as follows:
Find $f(t)$ (positive and independent of $n$) such that
$$n^n = \int_0^\infty t^n f(t) \,dt,\quad \forall\,n$$
That is, find $f(t)$ such that $\{n^n\}$ is the sequence of moments of f(t)dt.
This is a particular instance of the Stieltjes moment problem.
There is criterion (due to Carleman) which guarantees that such
a solution is unique, but I don't know an explicit form for $f(t)$. 
