# $(\frac1{n^{\sqrt n}})^{\frac1n}=(n^{-\frac{\sqrt n}{n}})^{}$

I'm a little bit confused by this one. Is this correct?

$$\left(\frac1{n^{\sqrt n}}\right)^{\frac1n}=\left(n^{-\frac{\sqrt n}{n}}\right)^{}=\sqrt{n^{-\frac1n}}$$

${}{}{}{}{}{}{}{}{}{}{}{}$

Edit: Is it okay that I changed the question a little bit ?

-
Maybe this is a little bit stupid question, shall I delete it ? – Kasper Feb 15 '13 at 17:17
There are no stupide questions :) – Dominic Michaelis Feb 15 '13 at 17:18

For your original question: Yes, the statement equality holds:

$$\left(\frac{1}{n^{\sqrt{n}}}\right)^{\Large\frac{1}{n}} = \left(n^{-\sqrt{n}}\right)^{\Large\frac{1}{n}}= \left(n^{-\Large\frac{\sqrt{n}}{n}}\right)$$

For your edited/added (second) (in)equality: Please note that, in fact $$\left(n^{-\Large\frac{\sqrt n}{n}}\right)^{}\neq \sqrt{n^{-\Large\frac1n}}$$ What we do have is this: $\quad\displaystyle \left(n^{-\Large\frac{\sqrt{n}}{n}}\right) = \left(\large n^{\large-\sqrt{\Large\frac 1n}}\right)$

-
Yes, definitely did clear things up, thanks :) – Kasper Feb 15 '13 at 18:52
+1 nice amWhy. Greeeeat.. – Babak S. Feb 15 '13 at 19:34

Yeah this is correct because $$\left(\frac{1}{n^{\sqrt{n}}}\right)^\frac{1}{n} = (n^{-\sqrt{n}})^\frac{1}{n}= (n^{-\frac{\sqrt{n}}{n}})$$

-

Recall: $$\left(\frac{1}{b^n}\right) = b^{-n} \tag{1}$$ $$\left(a^n\right)^m = a^{nm}\tag{2}$$

Now we have: $$\left(\frac{1}{n^{\sqrt{n}}}\right)^\frac{1}{n} = \left(n^{-\sqrt{n}}\right)^\frac{1}{n} \tag{applying equation (1)}$$

$$\left(n^{-\sqrt{n}}\right)^\frac{1}{n} = n^{-\sqrt{n}\cdot\frac{1}{n}} \tag{applying equation (2)}$$

Simplifying: $$n^{-\sqrt{n}\cdot\frac{1}{n}} = n^\frac{-\sqrt{n}}{n}$$

Therefore, yes, the statement in your question is correct.

-

Yes. Seems to be exactly right.

-