By definition

$$E.f(x)=f(x+h)---- (1)$$

$\triangle f(x)=f(x+h)-f(x)\\ \triangle f(x)=Ef(x)-f(x)\\ \triangle f(x)=(E-1)f(x)\\ \therefore \triangle =E-1 \\ \therefore E=1+\triangle ----- (2)\\ Also \space Ef(x)=f(x+h)\\ = f(x)+hf'(x)+\dfrac {h^2}{2!} f'(x) +\dfrac {h^3}{3!}f''(x) + .... \text{ Taylor series }\\ = f(x)+hDf(x)+\dfrac {h^2}{2!}D^2f(x)+\dfrac {h^3}{3!}D^3 f(x) ... where \space D^nf(x)=f^n(x)\\ =\Bigg[1+\dfrac {(hD)^1}{1!}+\dfrac {(hD)^2}{2!} + \dfrac {(hD)^3}{3!} + ....\Bigg] f(x)\\ \therefore Ef(x)=e^{hd}f(x)\\ \therefore E=e^{hd}----- (3)\\ From (2) \space and (3) \\ E=1+\triangle =e^{hD}$