Evaluate this number. Is it close to 1 or 0?

Solution 1:

${m \choose n}=\frac {m(m-1)(m-2)...(m-n+1)}{n!}\lt \frac {m^n}{n!}$ so your fraction is less than $\frac 1{n!}$. For $n \approx 10^{14}$ this is very small. This is not the probability that two decks out of $10^{14}$ match. See the generalized birthday problem for that. Very roughly, you will have a reasonable chance of a match when you have a number of samples that is the square root of the size of the sample space. With $52! \approx 10^{68}$ you need about $10^{34}$ samples to have a reasonable chance of a match. Wikipedia does a better job with the constants.

Solution 2:

$$x = \frac{\binom mn}{m^n}=\frac{\Gamma (m+1)}{m^n\,\Gamma (n+1) \Gamma (m+1-n)}$$ Take logarithms and use Stirling approximation for large $m$ since $52! \sim 8\times 10^{67}$ $$\log(x)=-\log (n!)-\frac{(n-1) n}{2 m}+O\left(\frac{1}{m^2}\right)\implies x \sim \frac 1{n!}$$

Edit (after the edit of the question)

$$x = \frac{n!\,\binom mn}{m^n}=\frac1{m^{n}}\frac{ \Gamma (m+1)}{\Gamma (m-n+1)}$$ Do the same as above to obtain

$$\log(x)=-\frac{(n-1) n}{2 m}-\frac{(n-1) n (2 n-1)}{12 m^2}+O\left(\frac{1}{m^3}\right)\tag 1$$ $$x=e^{\log(x)}=1-\frac{(n-1) n}{2 m}+O\left(\frac{1}{m^2}\right)$$ and this $n$ is very large, then $$x \sim 1-\frac {n^2}{2m}$$ which was already given by @Brian Tung in comments.

Using your numbers $\log(x) \sim -6.199000\times 10^{-41}$ while $(1)$ gives the same number if you accept an absolute difference of $1.588\times 10^{-149}$.