Is there a Definite Integral Representation for $n^n$?

This may not be the line of thinking you are after, but it does give the integral you desire.

$$n^n=\int_0^n nx^{n-1}\ dx$$

$$\frac{\Gamma (\alpha)}{s^{\alpha}}=\int_{0}^{\infty }t^{\alpha-1}e^{-st}dt\tag{1}$$

$$\frac{\Gamma (s)}{s^{s}}=\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{2}$$

$$s^{-s}=\frac{1}{\Gamma (s)}\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{3}$$

$$s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}\tag{4}$$

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 :

$$n E_{1-n}(\frac{1}{n})= \int_{1}^{\infty} e^{-\frac{t^{\frac{1}{n}}}{n}} dt\tag{4}$$