Denote by $I$ your integral. Then,
$I = e^N \int_0^1 {x^{N - 1} e^{ - xN} \,{\rm d}x} = \frac{{\Gamma (N)e^N }}{{N^N }}\int_0^N {\frac{{x^{N - 1} e^{ - x} }}{{\Gamma (N)}}\,{\rm d}x}.$
Now, if $X_1,\ldots,X_N$ are independent and identically distributed exponential(1) random variables, then their sum $X_1 + \cdots + X_N$ has gamma density $x^{N-1}e^{-x}/\Gamma(N)$, $x>0$. Thus, the last integral above is equal to ${\rm P}(X_1 + \cdots + X_N \leq N)$, or equivalently to ${\rm P}(X_1 + \cdots + X_N - N \leq 0)$. By the central limit theorem, ${\rm P}(X_1 + \cdots + X_N - N \leq 0) \to 1/2$ as $N \to \infty$ (since the $X_i$ have expectation equal to $1$). So, using $N^N \sim N!e^N /\sqrt {2\pi N} $, we get
$ I \sim \sqrt {\frac{\pi }{2}} \frac{1}{{\sqrt N }}.$