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Note that:

${14}^{{14}^{14} \times \frac{1}{14}}={14}^{{14}^{14} \times {14}^{-1}}=\boxed{{14}^{{14}^{13}}}$

One property of roots is that $\sqrt[n]{m}=m^{\frac{1}{n}}$

So $\sqrt[14]{14^{14^{14}}}=({14^{14^{14}}})^{\frac{1}{14}}=\boxed{{14^{14^{13}}}}$

From the property of exponents : If $k = m \times n$ , then $(a^{m})^{n} = a^{k}$ .

In this case of this problem : let the variables $\ \ \ a = 14 \ \ \ , \ \ \ k = 14^{14} \ \ \ , \ \ \ m = 14^{13} \ \$ then $\ \ \ n = 14^{1}$ . $\therefore \ \ a^{k} = 14^{14^{14}} = (a^{m})^{n} = (14^{14^{13}})^{14}$ $\therefore \ \ \sqrt[14]{14^{14^{14}}} = \sqrt[14]{(14^{14^{13}})^{14}} = \boxed{14^{14^{13}}}$

Problem: What is the 14th root of ${14}^{{14}^{14}}$ ?

Solution: We wish to solve this expression. To make this easier to understand, we could let $x={14}^{14}$ . Then, the expression can be written as $\sqrt[14]{{14}^{x}}$ . Next, recall that a radical can be expressed as a fractional exponent. So, our expression can be written as $({{14}^{x}})^{\frac{1}{14}}$ . Using the power rule, we can see that it simplifies to ${14}^{\frac{x}{14}}$ . Substituting the value of $x$ into the equation and simplifying it, we get the answer ${14}^{{14}^{13}}$ . $\square$

$14^{14}=14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14$

$14^{(14^{14})}=14^{(14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14)}$

The 14th root of $14^{(14^{14})}=(14^{(14^{14})})^{1/14}=14^{(14^{14})1/14}$

$=14^{(14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14)/14}$

$=14^{14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times\frac{14}{14}}$

$=14^{14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times14\times1}$

$=\boxed{14^{14^{13}}}$