Limit of $(n-k)! \cdot n^k$ as $n$ approaches infinity Is it true that $(n-k)! \cdot n^k$ tends to $n!$ as $n \to \infty$?
I think it is correct but can't think of a satisfying proof.

Sterling's approximation says $\ln (x!) = x \ln x - x + \frac12\ln(2\pi x) + O(\frac1x)$.
$$\ln \left((n - k)! n^k\right) = k \ln n + \ln \left((n - k)!\right)$$
$$ = k \ln n + (n - k)\ln(n - k) - (n - k) + \frac12 \ln (2 \pi (n - k)) + O(\frac1{n-k})$$
Since $\ln (n - k) = \ln\left(n(1 - \frac{k}n)\right) = \ln n + \ln (1 - \frac{k}n)$
$$ = k \ln n + (n - k)(\ln n + \ln(1 - \frac{k}n)) - n + k + \frac12 \ln (2 \pi (n - k)) + O(\frac1{n-k})$$
$$ = k \ln n + n \ln n - k \ln n + (n - k)\ln(1 - \frac{k}n) - n + k + \frac12 \left(\ln (2 \pi) + \ln n + \ln(1 - \frac{k}n)\right)  + O(\frac1{n-k})$$
$$ = n \ln n + (n - k + \frac12)\ln(1 - \frac{k}n) + - n + k \frac12 \left(\ln (2 \pi) + \ln n \right) + O(\frac1{n-k})$$
Since for $x \ll 1$ we have $\ln(1 - x) = -x - O(x^2)$
$$ = n \ln n + (n - k + \frac12)\left(-\frac{k}n - O(\frac{k^2}{n^2})\right) - n + k  \frac12\ln(2\pi n) +  O(\frac1{n-k})$$
$$ = n \ln n - k + \frac{k^2}n  - \frac{k}{2n} - O(\frac{k^2}{n})- n + k +\frac12\ln(2\pi n) +  O(\frac1{n-k})$$
$$ = n \ln n - n + \frac12\ln(2\pi n) + O(\frac{k^2}n) $$
$$ = \ln(n!) + O(\frac1n) + O(\frac{k^2}n) = \ln(n!) + O(\frac{k^2}n)$$
So $(n - k)!n^k$ is approximately equal to $n!$ within a factor of about $e^{\frac{k^2}n}$.  If you add in more terms of Sterling's approximation you might get a better big-O term.
