I have a simple question, is $n^{n\cdot n}=(n^n)^n$? I believe it does, because, for example $(n^2)^2 = n^{2\cdot 2}$. Also $1^1 = 1\cdot1, 2^2 = 2\cdot2, 3^3 = 3\cdot3, 4^4 = 4\cdot 4$, so I suppose $n^n = n\cdot n$, am i right?
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1yes............................ – Will Jagy Jan 20 '19 at 04:17
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But $3*3=9\not=27=3^3$. – Jens Schwaiger Jan 20 '19 at 04:19
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@Jens Schwaiger oh you are right, thank you – we_mor Jan 20 '19 at 04:22
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A law of exponents says that $$ (a^b)^c = a^{bc}. $$ In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
Randall
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Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication. – Jan 20 '19 at 05:25
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I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing. – Randall Jan 20 '19 at 05:34
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Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning. – Jan 20 '19 at 05:38