Refer to the definition of exponential function, for example from Wikipedia :
The functions such as $f(x)=e^{\sqrt x}$ and $e^{x^2 +1}$ are composite functions on the form $f(x)=e^{g(x)}$. They are not exponential functions according to the above definition, except if $g(x)=cx+d$ with constant $c,d$ .
They are "composite exponential functions" (I am not sure of the correctness of the translation. Please correct it if necessary).
For example in case of $g(x)=\ln(x)$ the fonction $f(x)=e^{g(x)}=x$ is not an exponential function.
If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$, you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate.
Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$. Now, compute the ratio between $f(x+1)$ and $f(x)$: $$
\frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, .
$$
As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$, thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$. Now analyze the ratio between $b(x+1)$ and $b(x)$. By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$, so the ratio between $b(x+1)$ and $b(x)$ is always $b$. However, we stated already that $b>1$. We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$.
Note that I just used terminology like approach and really big because a $7$th grader would not know of limits, so don't nitpick that.
Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$-axis.
If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$, then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.
Best Answer
See that $(5^2)^4=5^{2\cdot 4}=5^{8}\not=5^{2^4}=5^{(2^4)}=5^{16}$ according your link on Wikipedia.