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I am trying to bound values of very large factorials (N>500) without using Sterling's formula.

For large N, I found that n!>$100^n$ is a relatively good lower bound for the factorial. After trying for a while, I can't find a higher bound for factorials that is actually relatively close to the true value of the factorial.

Any suggestions for tighter lower and higher bounds are appreciated. Thanks

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    $100^n$ is just a lower bound for sufficiently large $n$. So ist $1000^n$ and in general $N^n$. Such functions are not growing fast enough. You must look at $n^n$ or better $\left(\frac ne\right)^n$ but this comes very close to Striling's formula. – 2017-02-23
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    What is the reason for avoiding Stirling's approximation? – 2017-02-23
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    For very large $n$, for example $n=10^{10^{100}}$, $n^n$ and $n!$ are virtually indistinguishable. – 2017-02-23
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    I haven't really learnt Stirling's approximation properly, and this is a sub-part to a paper that I was writing for school and we are not allowed to use math that has not been taught – 2017-02-23
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    For $n\ge 269$, you have $100^n – 2017-02-23
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    @Sumant If you want an approximation that is asymptotically correct, you can't avoid Stirling. But there are better elementary bounds than the one you give above. – 2017-02-23
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    If you are happy with the approximation of $100^n$ for lets say $n=500$, then you can choose $n!\approx n^n$ for every $n$. This will be a relative good approximation in the same sense that $100^{500} \approx 500!$ is relatively good. – 2017-02-23
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    I have to bound the expression $$\frac{(5000!)}{((1250!)^4)*[(4)^$(5000)$]}$$ is there any way to do this using stirling approximation? – 2017-02-23
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    sorry the denominator should be (1250)^4*(4^5000) – 2017-02-23
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    Try Ramanujan approximation for n!, you can find it here: https://www.johndcook.com/blog/2012/09/25/ramanujans-factorial-approximation/ – 2017-02-23

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