More understanding!

$1)\large{ \left( { n }^{ n } \right) }^{ { n }^{ n } }={ n }^{ { n\cdot n }^{ n } }={ n }^{ { n }^{ n+1 } }$ $2)\large{ \left( { n }^{ { n }^{ n } } \right) }^{ n }={ n }^{ { n }^{ n }\cdot n }={ n }^{ { n }^{ n+1 } }$ Also, statement above is true .

Let us prove the truth by putting $log$ on both sides of the equation.

$\therefore log (n^{n})^{n^n} = log \big( n^{n^n} \big) ^n$ .

Hence, $n^n \: log (n^n) = n \: log \big( n^{n^n} \big)$ .

Therefore, we have $n^n \times n \: log \, (n) = n \times n^n log \: n$ .

Here, both the sides of the equation are proved to be equal. Hence, for all positive integer n, the identity is $\boxed{true}$ .

Let $x = n$ , $y = n$ , and $z = n^n$ . Then the statement becomes $(x^y)^z = (x^z)^y$ , which is true for all $x, y, z$ (both expressions simplify to $x^{yz}$ ).

I claim that this is $\boxed{\text{true}}$ . We have $\begin{aligned}(n^n)^{n^n}=(n^n)^{n^1\times n^n}=(n^n)^{n+1}\\ \left(n^{n^n} \right)^n=n^{n^n\times n^1} = (n^n)^{n+1}\end{aligned}$ They are the same, so the answer is true.