Upper bound of $(1-1/a)^{b-1}$

Question

I have seen in a proof that for $a,b$ positive integers and for $b\geq a+1$ it is true that $$(1-1/a)^{b-1}\leq e^{-1}$$ but this is not clear to me by the exponential series and the obvious upper bound is $1$.

Answer 1

Hint Here is one way - note $e^x$ is convex and hence stays above its tangent at $x=0$, so $e^x \ge 1+x$. Now set $x =-\frac1a$ and simplify.