Answer
See the proof below.
Work Step by Step
In an adiabatic process, we know that the heat exchange is zero, $Q=0$.
And we know that the change in the thermal energy of this system is given by
$$\Delta E_{th}=W+Q$$
Hence,
$$\Delta E_{th}=W+0$$
where $\Delta E_{th}=nC_{\rm V}\Delta T$
$$W_s=-nC_{\rm V}(T_f-T_i)\tag1$$
Recall the ideal gas law,
$$PV=nRT$$
Hence,
$$T=\dfrac{PV}{nR}$$
Plugging into (1);
$$W_s= -\color{red}{\bf\not} nC_{\rm V}\left(\dfrac{P_fV_f}{ \color{red}{\bf\not} nR}-\dfrac{P_iV_i}{ \color{red}{\bf\not} nR}\right)$$
$$W_s=- \dfrac{C_{\rm V}}{R}\left(P_fV_f-P_iV_i\right)$$
Recalling that $$R=C_{\rm P}-C_{\rm V}$$
$$W_s=- \dfrac{C_{\rm V}}{C_{\rm P}-C_{\rm V}}\left(P_fV_f-P_iV_i\right)$$
$$W_s=- \dfrac{\frac{C_{\rm V}}{C_{\rm P}}}{\frac{C_{\rm P}}{C_{\rm P}}-\frac{C_{\rm V}}{C_{\rm P}}}\left(P_fV_f-P_iV_i\right)$$
Recalling that $\gamma=C_{\rm P}/C_{\rm V}$
$$W_s=- \dfrac{\frac{1}{\gamma}}{1-\frac{1}{\gamma}}\left(P_fV_f-P_iV_i\right)$$
$$W_s=- \dfrac{\frac{1}{ \color{red}{\bf\not} \gamma}}{ \frac{\gamma-1}{ \color{red}{\bf\not} \gamma}}\left(P_fV_f-P_iV_i\right)$$
$$W_s=- \dfrac{1}{ \gamma-1}\left(P_fV_f-P_iV_i\right)$$
$$\boxed{W_s= \dfrac{P_fV_f-P_iV_i }{1- \gamma } }$$
