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Let $a \in R$, such as $a>1$. How to find smallest natural number $N$, such that $a^N < N!$?

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    Stirling's formula. For $n>0$ we have $N!=(N/e)^N\sqrt {2\pi N}\;(1+d_N)$ where $0 – 2017-01-09
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    This formula gives nothing about $N$ – 2017-01-09

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In fact @Robert Israel has found a formula to the sequence.

A small check of the smallest $N$ for $a =1,2,\cdots 10$ gives us the sequence $A065027$ in the OEIS giving us the smallest $N >0$ such that $a^N

He writes that it appears that $L(n) < a(n) - n e + \log{\sqrt{2 \pi n)}} < \frac {1}{2} $, where $L(n) = -\frac {1}{2} + o(1)$ , and $L(n) > -0.53$ for all $n$.

Hope it helps.

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    It is interesting idea, I will check it out – 2017-01-09
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by taking the logarithm on both sides we obtain $$N\ln(a)<\ln(N!)$$ from here we get $$\ln(a)<\frac{\ln(N!)}{N}$$ thus we get $$a

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    I think that the OP has a $!$ as a factorial,though I'm not sure. – 2017-01-09
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    yes ok i have corrected it – 2017-01-09
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    Even with an edit this doesn't really help. You can estimate an $a$ given an $N$, but OP is asking the reverse of that. – 2017-01-09
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    one can solve the reverse of this problem only by a numerical method – 2017-01-09