Large $n$ asymptotic of $\int_0^\infty \left( 1 + x/n\right)^{n-1} \exp(-x) \, \mathrm{d} x$ While thinking of 71432, I encountered the following integral:
$$
  \mathcal{I}_n  = \int_0^\infty \left( 1 + \frac{x}{n}\right)^{n-1} \mathrm{e}^{-x} \, \mathrm{d} x
$$
Eric's answer to the linked question implies that $\mathcal{I}_n \sim \sqrt{\frac{\pi n}{2}} + O(1)$.
How would one arrive at this asymptotic from the integral representation, without reducing the problem back to the sum ([added] i.e. expanding $(1+x/n)^{n-1}$ into series and integrating term-wise, reducing the problem back to the sum solve by Eric) ? 
Thanks for reading. 

I shifted the function by a unit since it won't effect the asymptotics and I'd like the global maximum to occur at $x=0$. 
$$ \mathcal{I}_n \sim \int^{\infty}_0 \left( 1 + \frac{x-1}{n} \right)^{n-1} e^{-(x-1) } dx $$
$$\left( 1 + \frac{ x-1}{n} \right)^{n-1} e^{-(x-1) } = e \left( 1 - \frac{1}{n} \right)^{n-1} \left( 1 - \frac{x^2}{2(n-1)} + \cdots \right)$$
$$ \approx e \left(1 - \frac{1}{n} \right)^{n-1} \exp\left(\frac{-x^2}{2(n-1)} \right) $$
so $$ \mathcal{I}_n \sim e\left( 1 -\frac{1}{n}\right)^{n-1} \int^{\infty}_0 \exp\left( \frac{-x^2}{2(n-1)} \right) dx $$
$$ = e\left( 1 -\frac{1}{n}\right)^{n-1} \sqrt{\pi(n-1)/2} \sim \sqrt{\pi n/2}$$
