How can we show $\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $ for all $n \in \mathbb{N}$, $0 \leq c \leq x \leq 1$, $c \neq 1$? 
How can we show $$\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $$ for all $n \in \mathbb{N}$, $0 \leq c \leq x 
 \leq 1, c \neq 1$?

The context is this probability problem, but of course this problem might be of independent interest to inequality enthusiasts.
In my attempt, I have graphed the inequality on Desmos. We might note that the derivative changes sign in the range $c < x < 1$ so it may be unlikely differentiation would be of help.

If $x=c$ or $n=1$ then inequality holds. So consider $x > c, n > 1$:
$$\frac{1-x^n}{1-c^n}+\left(\frac{x-c}{1-c}\right)^n\leq 1\Leftrightarrow
1-x^n+(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq 1-c^n\Leftrightarrow$$
$$(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq x^n-c^n\Leftrightarrow
\frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
Consider $f(x)=\frac{x^n-c^n}{(x-c)^n}$, then $$f'(x)=\frac{n x^{n-1}}{(x-c)^n}-\frac{n(x^n-c^n)}{(x-c)^{n+1}}=\frac{nc(c^{n-1}-x^{n-1})}{(x-c)^{n+1}}<0$$
$$f'(x)<0, x\leq 1 \Rightarrow f(1)\leq f(x)\Rightarrow \frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
