Prove that ${^{n}}{P}_{r} = {^{\left(n - 1\right)}}{P}_{r} + r \cdot {^{\left(n - 1\right)}}{P}_{\left(r - 1\right)}$
By definition,
${^{n}}{P}_{r} = \dfrac{n!}{\left(n - r\right)!}$ $\;\;\; \cdots \; (1)$
${^{\left(n - 1\right)}}{P}_{r} = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!}$ $\;\;\; \cdots \; (2)$
${^{\left(n - 1\right)}}{P}_{\left(r - 1\right)} = \dfrac{\left(n - 1\right)!}{\left(n - r\right)!}$ $\;\;\; \cdots \; (3)$
$\therefore \;$ We have from equations $(2)$ and $(3)$,
$\begin{aligned}
{^{\left(n - 1\right)}}{P}_{r} + r \cdot {^{\left(n - 1\right)}}{P}_{\left(r - 1\right)} & = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} + r \cdot \dfrac{\left(n - 1\right)!}{\left(n - r\right)!} \\\\
& = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} + \dfrac{r \cdot \left(n - 1\right)!}{\left(n - r\right) \left(n - r - 1\right)!} \\\\
& = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} \left[1 + \dfrac{r}{n - r}\right] \\\\
& = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} \left[\dfrac{n}{n - r}\right] \\\\
& = \dfrac{n!}{\left(n - r\right)!} \\\\
& = {^{n}}{P}_{r} \;\;\;\; \left[\text{from equation (1)}\right]
\end{aligned}$
Hence proved.