Real Analysis question on sequences (Hint needed!!!) Given the number $\alpha > 1$ , define the sequence an where $a_0 = 1$ and $a_{n+1} = (\alpha \times a_{n})^{\frac{1}{4}}$ for $ n \geq 0 $. Prove: 

If $a_{n}^{3}< \alpha $(as is true when n = 0), then $a_{n} < a_{n+1} $and $a_{n+1}^3 < \alpha $. Conclude that the sequence is increasing and bounded. Show that the cube of the limit is $ \alpha$ . (Thus we have constructed $\alpha^\frac{1}{3}$.)

Can someone give me a hint to start it?

HINTS: If $$a_{n+1}=(\alpha\,a_n)^{1/4}$$ then $$\frac{a_{n+1}}{a_n}=\left(\frac{\alpha}{a_n^3}\right)^{1/4}$$ Can you show that this is greater than $1$?
Also, $a_n^3<\alpha\iff a_n<\alpha^{1/3}$ so we have $$a_{n+1}<(\alpha\cdot\alpha^{1/3})^{1/4}=(\alpha^{4/3})^{1/4}$$ These should be helpful.
