Answer
Let $x=100$. Then $2^{x}=2^{100}=(2^{10})^{10}$ and $x^{10}=100^{10}$. But since $2^{10}=1024\gt100$, it must be that $(2^{10})^{10}=2^{100}\gt100^{10}$, so $x=100$ satisfies the hypotheses. We conclude that there are real numbers $x$ such that $x\gt1$ and $2^{x}\gt$$x^{10}$.
Work Step by Step
This proof is a constructive proof of existence, whereby we show that something exists by finding a specific example. For more on this method of proof, see the section entitled "Proving Existential Statements" beginning on page 148, especially example 4.1.3.
Note that the number $100$ was chosen in this example because it is easy to show by hand that it satisfies the hypotheses and conclusion, since $2^{100}$ may then be written as $(2^{10})^{10}$. However, many other numbers would have worked just as well.
