Elegant proof of $e^{nf}=\cosh(n)+\sinh(n)f$ where $f^2=1$ The following transition comes up in various physics texts, often without explanation, so I was wondering if there is a nice way to prove it:
Given $n$ and $f$ such that $f^2=1$, we have 
$$e^{nf}=\cosh(n)+\sinh(n)f$$
I managed to prove it through series expansion, but it is far from elegant.
I tried to make sense of a different way of viewing it, $$e^{nf}=\frac {e^n+e^{-n}}2+\frac {e^n-e^{-n}}2f$$
but I still don't see it.
Any idea?
 A: $$\cosh(n)=\frac{e^n+e^{-n}}{2}$$
$$\sinh(n)=\frac{e^n-e^{-n}}{2}$$
There is only 2 possible $f$ value, so it's easy to check for both, Let's start with $f=1$:
$$\cosh(n)+\sinh(n)=\frac{e^n+e^{-n}}{2}+\frac{e^n-e^{-n}}{2}=2\frac{e^n}{2}+\frac{e^{-n}}{2}-\frac{e^{-n}}{2}=2\frac{e^n}{2}=e^{n}=e^{fn}=\cosh(n)+\sinh(n)f$$
Now $f=-1$:
$$\cosh(n)-\sinh(n)=\frac{e^n+e^{-n}}{2}-\frac{e^n-e^{-n}}{2}=2\frac{e^{-n}}{2}+\frac{e^{n}}{2}-\frac{e^{n}}{2}=2\frac{e^{-n}}{2}=e^{-n}=e^{fn}=\cosh(n)+\sinh(n)f$$
Q.E.D.
Edit: I've seen the comments about the possibility of a different $f$, so I'm trying to prove it a different way.
Let's examine the ratio of the two side. If it's derivate is $0$, then their ratio is constant (assuming associativity, distributivity, commutativity):
$$\frac{\mathrm{d}}{\mathrm{d}n}\bigg(\frac{\cosh(n)+f\sinh(n)}{e^{fn}}\bigg)=\frac{\mathrm{d}}{\mathrm{d}n}\bigg((\cosh(n)+f\sinh(n))e^{-fn}\bigg)=-fe^{-fn}(\cosh(n)+f\sinh(n))+e^{-fn}(\sinh(n)+f\cosh(n))=e^{-fn}(-f\cosh(n)+f\cosh(n)-f^2\sinh(n)+\sinh(n))=e^{-fn}\sinh(n)(1-f^2)=e^{-fn}\sinh(n)(0)=0$$
Now substitute $n=0$ into the fraction to get their ratio:
$$\frac{\cosh(0)+f\sinh(0)}{e^{f0}}=\frac{1+0f}{1}=1$$
So the $2$ sides are equal.
Q.E.D.
