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Why is $a^{b^{c^d}}$ not equal to ${(a^{b^c})}^d$ (for positive n)?

For example, WolframAlpha seems to say that $2^{2^{2^n}}$ is not equal to $16^n$.

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    $\begingroup$ Possible duplicate of What is the solution of $2^{2^{2^2}}$? $\endgroup$ – Ross Millikan Nov 23 '16 at 19:55
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    $\begingroup$ @Did How is 2^2^2^n ambiguous? It's clearly 2^(2^(2^n)) $\endgroup$ – Max Nov 23 '16 at 19:58
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    $\begingroup$ @Max And what happens if somebody insists that it is "clearly" ((2^2)^2)^n? You call the Pope for their advice? $\endgroup$ – Did Nov 23 '16 at 19:59
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    $\begingroup$ @Max Not every author uses this convention (which seems to be originating from a programming context rather than from mathematics). Hence: ambiguous. Hence: to be avoided. $\endgroup$ – Did Nov 23 '16 at 20:02
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    $\begingroup$ @Max, how is this priority rules? Exponentiation has higher priority then exponentiation? This is not a case of $2+2^2$ for priority to decide, this is a question of left/right associativity. I do agree that right associativity is usual here (I've never seen someone treating it as left associative) but this is not obvious, it is a convention (differing from convention for subtraction, for example). $\endgroup$ – Ennar Nov 23 '16 at 20:08
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It is purely a matter of notational convention that $a^{b^c}=a^{(b^c)}$ rather than $(a^b)^c$, but it's a convention that makes sense: There would be no point in using the notation for the second convention, since it would be easier to simply write $a^{bc}$ -- and likewise for $a^{b^{c^d}}$ if the notation meant $a^{b^{c^d}}=(a^{b^c})^d=((a^b)^c)^d=a^{bcd}$. One reason for adopting it as a convention is that the unambiguous notation

$$a^{\left(b^{\left(c^d\right)}\right)}$$

takes up way more room.

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The problem is here what you mean by

a^b^c

For example: $$ (2^3)^2 = 8^2 = 64 \\ 2^{(3^2)} = 2^9 = 512 $$ Or $$ (2^{(2^2)})^3 = 16^3 = 4096\\ (2^2)^{(2^3)} = 4^8 = 65536 $$ So there is an example with $n=3$.

You want to use paratheses to make it clear what you want to do.

In fact recall the rule that $$ (a^b)^c = a^{b\cdot c} $$

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The operation ^ is not associative, i.e. in general $a^{(b^c)}\neq (a^b)^c$. The usual convention is $a^{b^c}=a^{(b^c)}$.

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Although I think Barry Cipra's answer should be the canonical one, I think there is also an argument against the OP's proposal on typographical grounds: if the intent of $a^{b^c}$ was to convey $(a^b)^c$, then there would be no justification for making the $c$ smaller, since it is then a top-level exponent and not a superscript within a superscript: it should look like ${a^b}^c$ rather than $a^{b^c}$. The fact that the $c$ is intentionally typeset smaller than $b$ establishes the intent that the $c$ is an exponent within a term that is already in smaller type size, namely the exponent of $a$.

Of course, semantics need not always follow syntactic structure literally. For instance, there is a practical limit to how small one can make subscripts, so in a tall tower of exponents we might see exponents at different heights with the same type size. But I can see no reason for exponents at the same height to have different sizes.

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The mistake people make usually is they go from down to up rather than coming from up to down. $${16^n}$$ is nothing but $${(2^n)(2^n)(2^n)(2^n)}$$.

On the other hand, While solving for $$2^{2^{2^n}}$$ You will have to come down by giving the upper powers to the 2's which are lying below.

For example $$2^{2^{2^2}}=(2^{16})$$

If we use your terminology it will be 256 which is false.

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  • $\begingroup$ What does "terminology" mean in the last sentence? $\endgroup$ – Erick Wong Nov 23 '16 at 21:08
  • $\begingroup$ @ErickWong, I just wanted to say that if the OP will use the method he is asking, the answer will be wrong. $\endgroup$ – Vidyanshu Mishra Nov 23 '16 at 21:11
  • $\begingroup$ Perhaps "notation" makes more sense in the last line. $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 20:42
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Note: $a^{b^{c^d}} = a^{b^{(c^d)}} \neq {(a^{b^c})}^d$

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