I hope this has not been asked already.
while reading a material on upper bounds of polynomials (time complexity) they compared growth of various polynomial functions (like $N\log N$, $N^2$, $N^3$, $2^N$ etc.)
While comparing ($k^N$) and $N!$ I had a doubt on their crossover-point, i.e, when one function exceeds the other.
So, If $k$ is a constant $\lt N$
For what minimum value of $N$ does the function $k^N$ cross/exceed $N!$ ? How to prove it?
On
You mean $n!$ exceeds $k^n$? This might be proved the following way:
$\frac{n!}{k^n}=\frac{k!}{k^k} \frac{(k+1) \cdots (n)}{k^{n-k}} \geq \frac{k!}{k^k} \frac{(k+1)^{n-k}}{k^{n-k}} = C (\frac{k+1}{k})^n$, where $C= \frac{k!}{k^k (\frac{k+1}{k})^k}= \frac{k!}{(k+1)^k}$. This function grow without bounds and one day will exceed $1$.
It is assumed that $k$ is an integer, but for any real $x$, you can find $k>x$ and proceed the same way for $x^n$.
By the Stirling formula, $n!\sim \left(\frac ne\right)^n$, so $n!>k^n$ at about the point where $\frac ne>k$ or $n> k\cdot e$.