[Math] Factorial grow faster than Exponential – permutation case

computational complexityfactorial

It is said that factorial grows faster than exponential, but in the case of permutation:

   permutation with repetition = n^n

   permutation without repetition = n!

And of course permutation with repetition has a bigger "space" than permutation without repetition which means it should grow faster.

I was looking for some convincing proof that shows this in an intuitive way.

Like for example displaying n! in a n^0.n^(m-1).n^(m-2)…n^1 and eventually shows that the sum of the exponential is < n, but did not succeed.

I am not sure if anyone can help me with the proof or maybe just help to explain the case against the "factorial grows faster than exponential" fact

Sorry for my wording, I have very poor mathematical backgrounds

Best Answer

The $n^n$ function is not an exponential as the basis varies.

To prove $n!=o(n^n)$, let's evaluate the ratio: $$\frac{n!}{n^n}=\frac1n\cdot\frac2n\cdots\frac{n-1}n\cdot \frac nn <\frac 1n$$ since each factor $\dfrac kn$ is less than $1$ if $k<n$. This inequality implies $\;\displaystyle\lim_{n\to\infty}\frac{n!}{n^n}=0$.

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