# $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means

Show that

$$\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}=+\infty$$

without using l'Hôpitals rule or any special means.

My first attempts were to use inequalities to approximate $n!$ and $n^n$ but I am really unsure what to do with $2^{n^3}$ and $5^{n^3}$. Any specific hints on how to transform this fraction?

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Hint: Try to estimate the denominator from above. For example $n! \le n^n \le (2^{n})^n = 2^{(n^2)}$. The basic idea is that $2^{(n^3)}$ is "so large" that you have much place to be generous with your estimations.
My results are always converging to 0 like $1/n!$ or $1/(5^{n^2}-1)$. Do you know what I might have done wrong? – Christian Ivicevic Nov 5 '12 at 14:29