Answer
See the proof below.
Work Step by Step
In order to show that the formula holds, we must show that given a real number $x$, the formula is true for $n=1$ and if it is true for $n=k$ then it is true for $n=k+1$.
Let $x$ be a fixed real number.
$\textbf{1.}$ The formula holds for $n=1$ and $x$ since
$$\left|x^1\right|=\left|x\right|=\left|x\right|^1.$$
$\textbf{2.}$ Suppose the formula holds for $n=k$ and $x$ that is, suppose that $$\left|x^k\right|=\left|x\right|^k.$$
Observe that $\left|x^{k+1}\right|=\left|x^k x\right|$.
Since $\left|ab\right|=\left|a\right|\left|b\right|$ for any pair of real numbers $a$ and $b$, it follows that
$$\left|x^k x\right|=\left|x^k\right|\left|x\right|.$$
Now, since $\left|x^k\right|=\left|x\right|^k$ and $\left|x\right|=\left|x\right|^1$ we have that
$$\left|x^k\right|\left|x\right|=\left|x\right|^k\left|x\right|^1=\left|x\right|^{k+1}.$$
Thus, $\left|x^{k+1}\right|=\left|x\right|^{k+1}$ and therefore, the formula holds for $n=k+1$ and for $x$ whenever it is true for $n=k$ and for $x$.
Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$ and $x$.
Since $x$ was an arbitrary real number, we have that the formula holds for every positive integer $n$ and for every real number $x$.
