$x^{1} \times x^{-1} = 1$

$x \times x^{-1} = x \times \frac{1}{x}$

For $x=0$ , $\frac{1}{0}$ is indetermiante and not one. Therefore the equation above does not satisfy all real values of $x$ .

Note: $x^{1} \times x^{-1} = x^{1+(-1)} = x^{0}$ , but $0^{0}$ remains indeterminate.

The answer is $\color{#D61F06} \boxed{\text{false}}$ .

When $x = 0,$

$0^1 \cdot 0^{-1} \text{is undefined}.$