How to prove the accuracy of this equation?

Question

How to prove the accuracy of this equation?

$$(\sinh x + \cosh x)^{n} = \sinh nx + \cosh nx$$

I already know:

$$\sinh x= \frac{e^{x} - e^{-x}}{2}$$ And $$\cosh x= \frac{e^{x} + e^{-x}}{2}$$ But I failed to prove it, since I don't see any point to start from.

Answer 1

Thanks to the hints of Blitzer and David Quinn found a way.

Simplified LHS: $$({\frac{e^{x} - e^{-x} + e^{x} + e^{-x}}{2}})^{n} = $$ $$(\frac{2e^{x}}{2})^{n} = e^{nx}$$

Simplified RHS: $${\frac{e^{nx} - e^{-nx} + e^{nx} + e^{-nx}}{2}} = $$ $$\frac{2e^{nx}}{2} = e^{nx}$$

LHS = RHS, so proved. Thanks.

Answer 2

$$(\sinh (x) + \cosh (x))^{n} = \sinh (nx) + \cosh (nx)$$ Let $x=i\,y$ to make $$(\cos (y)+i \sin (y))^n=\cos (n y)+i \sin (n y)$$