[proofplan] We express the change in fitted value caused by deleting observation $i$ in terms of the full residual $e_i$. The deleted-residual identity converts $e_{i(i)} - e_i$ into the factor $e_i h_{ii}/(1-h_{ii})$. Dividing by the defining scale $s_{(i)}\sqrt{h_{ii}}$ for $\operatorname{DFFITS}_i$ then gives the desired product of the externally studentized residual and the leverage factor. [/proofplan] [step:Rewrite the change in fitted value using the two residual definitions] By definition, \begin{align*} e_i &= y_i - \hat{y}_i, \\ e_{i(i)} &= y_i - \hat{y}_{i(i)}. \end{align*} Subtracting the first identity from the second gives \begin{align*} e_{i(i)} - e_i &= (y_i - \hat{y}_{i(i)}) - (y_i - \hat{y}_i) \\ &= \hat{y}_i - \hat{y}_{i(i)}. \end{align*} Thus \begin{align*} \hat{y}_i - \hat{y}_{i(i)} = e_{i(i)} - e_i. \end{align*} [/step] [step:Use the deleted-residual identity to isolate the leverage factor] Using the assumed identity $e_{i(i)} = e_i/(1-h_{ii})$, we obtain \begin{align*} \hat{y}_i - \hat{y}_{i(i)} &= e_{i(i)} - e_i \\ &= \frac{e_i}{1-h_{ii}} - e_i \\ &= e_i\left(\frac{1}{1-h_{ii}} - 1\right) \\ &= e_i\left(\frac{h_{ii}}{1-h_{ii}}\right). \end{align*} Since $0<h_{ii}<1$, the denominator $1-h_{ii}$ is positive, so this expression is well-defined. [/step] [step:Divide by the DFFITS scale and identify the externally studentized residual] By the definition of $\operatorname{DFFITS}_i$ and the preceding identity, \begin{align*} \operatorname{DFFITS}_i &= \frac{\hat{y}_i - \hat{y}_{i(i)}}{s_{(i)}\sqrt{h_{ii}}} \\ &= \frac{e_i h_{ii}}{s_{(i)}(1-h_{ii})\sqrt{h_{ii}}} \\ &= \frac{e_i\sqrt{h_{ii}}}{s_{(i)}(1-h_{ii})}. \end{align*} Because $0<h_{ii}<1$ and $s_{(i)}>0$, all square roots and denominators in this computation are well-defined. Now factor the last expression as \begin{align*} \frac{e_i\sqrt{h_{ii}}}{s_{(i)}(1-h_{ii})} &= \frac{e_i}{s_{(i)}\sqrt{1-h_{ii}}}\sqrt{\frac{h_{ii}}{1-h_{ii}}} \\ &= t_i\sqrt{\frac{h_{ii}}{1-h_{ii}}}, \end{align*} where the final equality uses the definition of $t_i$. Therefore \begin{align*} \operatorname{DFFITS}_i = t_i\sqrt{\frac{h_{ii}}{1-h_{ii}}}, \end{align*} as claimed. [/step]