Finding asymptotic growth using the Laplace Method

I am asked to find the the asymptotic growth of the integral $$I(k,m)=\int_0^1{x^{nk}}{{(1-x)}^{nm}}dx$$ with $k,m>0$ fixed. How do I approach this solution? I know it means finding the growth as $n \to \infty$, but I am clueless on what to do. Does it have something to do with Stirling's approximation?

To actually derive the behavior for large $n$, we begin by rewriting in Laplace's integral form \begin{align} I(k,m)& = \int_0^1{x^{nk}}{{(1-x)}^{nm}}dx \\ & = \int_0^1 e^{n[k \ln x + m \ln(1-x)]} \ dx. \end{align} Define $\phi(k,m,x)$ such that the integral is $e^{n\phi}$. This is maximal when $\partial \phi/\partial x = 0$ and is given by $x_c = \frac{k}{k+m}$. You can check this is maximal. Next we return to the integral and expand $\phi$ to second order. \begin{align} I(k,m)& = \int_0^1{x^{nk}}{{(1-x)}^{nm}}dx \\ & = \int_0^1 e^{n[k \ln x + m \ln(1-x)]} \ dx \\ & \approx \int_{x_c - \epsilon}^{x_c + \epsilon} e^{n[\phi(x_c) - \frac{1}{2}|\phi''(x_c)|(x-x_c)^2]} \ dx \\ & = e^{n \phi(x_c)} \int_{-\epsilon}^{\epsilon} e^{-n|\phi''(x_c)|\frac{1}{2}u^2]} \ du \\ & = e^{n\phi(x_c)} \int_{-\sqrt{|\phi''(x_c)|n}}^{\sqrt{|\phi''(x_c)|n}} e^{-\frac{1}{2}s^2} \ \frac{ds}{\sqrt{|\phi''(x_c)|n}} \\ & \approx \frac{e^{-n\phi(x_c)}}{\sqrt{|\phi''(x_c)|n}} \int_{-\infty}^\infty e^{-\frac{1}{2}s^2} \ ds \\ & = e^{-n\phi(x_c)}\sqrt{\frac{2\pi}{|\phi''(x_c)|n}} \end{align} You can insert the values for $\phi$ evaluated at $x_c$ to check against the asymptotics.