Proof of exponent rule $a^{n/m}$ when $n$ and $m$ belongs in the reals Now I have not studied math for very long. I have just completed Calculus1, although my knowledge extends a bit outside of this. 
My question is, how can the transforms below be justified
$$ a^{n/m} \, = \, \left( a^{1/m} \right)^{n}  \, = \, \left( a^{n} \right)^{1/m} $$
when $n$ and $m$ can be any real number? 
Now the statement above is easily proven for the natural numbers. 
For an example we can can see that these rules makes sense when dealing with real numbers. 
$$ 8^{2/3} \, = \, \left( 8^{1/3} \right)^{2}  \, = \, \left( 8^{2} \right)^{1/3} $$
But I just can not wrap my head around how we can justify the below algebraic manipulations
$$  2^{\pi/e} \, = \, \left( 2^{1/e} \right)^{\pi}  \, = \, \left( 2^{\pi} \right)^{1/e} $$
As I have problems understanding what $2^{\pi} $ means. How would one go about multiplying a number by itself $\pi$ times?
Someone said that this exponent rule, could be explained by something called dedikins cuts, or something like that. Any insight to answer my silly questions is greatly appreciated. 


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*Is there a general proof of exponent rule $a^{n/m}$ when $n$ and $m$ belongs in the reals?

 A: The definition of $a^x$ can be done in steps.
For $x$ a nonnegative integer and $a\neq 0$, we define $a^x$ inductively: $a^0 = 1$, and $a^{n+1} = a^n\times a$. By mathematical induction this defines $a^m$ for all nonnegative integers $m$.
It is then a further exercise in induction to prove that $a^na^m = a^{n+m}$, $(ab)^n = a^nb^n$, and $(a^n)^m = a^{nm}$. For instance, fixing $n$, we prove that $a^na^0 = a^{n+0}$ (true, since $a^0=1$ by definition); assuming that $a^na^m = a^{n+m}$, then 
$$a^n(a^{m+1}) = a^n(a^ma) = (a^na^m)a = a^{n+m}a = a^{(n+m)+1} = a^{n+(m+1)}.$$
We can also define $0^n$ for any positive $n$ by letting $0^1 = 0$, $0^{n+1} = 0^n0$. 
Having defined $a^n$ for all nonzero real numbers $a$ and all nonnegative integers $n$, we extend the definition to negative integers as follows: if $m$ is a negative integer, $m=-n$ with $n$ a positive integer, then we define
$$a^m = a^{-n} = \frac{1}{a^n}.$$
It is now easy to check that $a^na^m = a^{n+m}$ still holds for any integers $n$ and $m$, that $(ab)^n = a^nb^n$, and that $(a^n)^m = a^{nm}$. We deal with the case of negative integer exponents by using the definition. For example, to show $a^{n+m}=a^na^m$, we note that if $n,m\geq 0$, then we have already proven it. If $n\gt 0$, $m\lt 0$, and $n+m\geq 0$, then
$$a^{-m}a^{n+m} = a^{n+m-m} = a^n$$
by the positive case, and multiplying both sides by $\frac{1}{a^{-m}} = a^m$ gives the equality. Similar arguments hold if $n+m\lt 0$. If both $n$ and $m$ are negative, then taking reciprocals reduces to the positive case.
Having defined $a^n$ for all $a\neq 0$ and all integers $n$, we extend the definition to rational numbers. In order to do that, we need to use some facts about the real numbers, and we need to restrict our choice of $a$: we can only do the following if $a\gt 0$.
Given a positive integer $q$, $a^{1/q}$ is defined to be the unique real number $r$ such that $r^q = a$. Such a number exists by the Intermediate Value Theorem (consider the function $f(x) = x^q$) and is unique (the function $f(x)=x^q$ is strictly increasing on $[0,\infty)$). 
Then given a rational number $p/q$ with $p$ and $q$ relatively prime, $q\gt 0$, we define $a^{p/q} = (a^{p})^{1/q}$.  If $n$ and $m$ are arbitrary integers, we define $a^{n/m}$ to be $a^{p/q}$, where $\frac{n}{m}=\frac{p}{q}$ and $p$ and $q$ are relatively prime, and $q\gt 0$. 
This definition restricts to the previous definition when $q=1$: by definition, $a^{p/1}= (a^p)^1$ is the unique real number $r$ such that $r^1 = a^p$; that is, $a^{p/1} = a^p$. 
Then one shows that this definition still satisfies the exponentiation rules: $a^{p/q}a^{r/s} = a^{(p/q)+(r/s)}$, $(ab)^{p/q} = a^{p/q}b^{p/q}$, and $(a^{p/q})^{r/s} = a^{pr/qs}$. 
For example, to show that $a^{(p/q)+(r/s)} = a^{p/q}a^{r/s}$, it suffices to show that $a^{p/q}a^{r/s}$ is the unique real number which, when raised to the $qs$th power, gives $a^{ps+qr}$, since $\frac{p}{q}+\frac{r}{s} = \frac{ps+qr}{qs}$. But this is merely an assertion about integer powers, so we can use the already established properties:
$$(a^{p/q}a^{r/s})^{qs} = (a^{p/q})^{qs} (a^{r/s})^{qs} = ( ( a^{p/q})^q)^s( ( a^{r/s})^s)^q.$$
Now, $a^{p/q} = b$ means that $b^q = a^p$. So
$$ ((a^{p/q})^q)^s = (a^p)^s = a^{ps}.$$
Similarly, $((a^{r/s})^s)^q = a^{rq}$. So
$$(a^{p/q}a^{r/s})^{qs} = a^{ps}a^{rq} = a^{ps+rq},$$
which proves that $a^{p/q}a^{r/s} = a^{(ps+rq)/qs} = a^{(p/q) + (r/s)}$.
Etc.
Having defined exponentiation for rational exponents, we extend it to exponentiation for any real number in one of several ways:
One way is as follows: Given a real number $r$, let $q_n$ be an increasing sequence of rational numbers such that $q_n\to r$. Then we define $a^r = \lim\limits_{n\to\infty}a^{q_n}$.
One needs to show that this is well-defined (the limits exists, and if $p_n$ is a different sequence of rationals converging to $r$, then the limit of $a^{p_n}$ equals the limit of $a^{q_n}$). This can be done; if you do that, then the fact that $(a^r)^s = a^{rs}$ follows by showing that if $p_n\to r$ and $q_m\to s$, then showing that
$$
\lim_{m\to\infty}\lim_{n\to\infty}(a^{p_n})^{q_m} = \lim_{m\to\infty}\lim_{n\to\infty}a^{p_nq_m}$$
will converge to the same thing as $a^{p_nq_n}$, and therefore to $a^{rs}$. 
Alternatively, one can define $a^r$ is the supremum of $a^{p/q}$ with $p/q$ rational numbers smaller than $r$. There is no issue of well-definedness here ($a^r$ exists and is unique), and then one just needs to show that we can express rationals smaller than $rs$ in terms of rationals smaller than $r$ and rationals smaller than $s$, and then use the properties of the exponentiation that we already know for rationals.
An entirely different approach is to define the exponential function $e^x$ using a Taylor Series:
$$e^a = \lim_{n\to\infty}\left(1 + \sum_{k=1}^n\frac{a^k}{k!}\right),$$
then showing that this has the "usual" properties, defining the logarithm as its inverse, and finally defining $a^b = e^{b\ln a}$. Then the rule $(a^b)^c = a^{bc}$ follows directly from the properties of the logarithm.
