Show $\left(1-\frac x k\right)^k<\left(1-\frac {x}{ k+1}\right)^{k+1}$ 
Show that $$\left(1-\frac{x}{k}\right)^k<\left(1-\frac{x}{k+1}\right)^{k+1}$$ for $x>0$, and $k \ge 1$, where $k$ is a whole number.

Is it possible to prove this? I can easily prove algebraically for $k=1$ case, but I am wondering if this is true in general. 
 A: Consider numbers $1$, $1-\dfrac{x}{k}$ ($k$-times) and apply $AM-GM$ then
$$\dfrac{1+(1-\dfrac{x}{k})+(1-\dfrac{x}{k})+(1-\dfrac{x}{k})+\cdots+(1-\dfrac{x}{k})}{k+1}\geq\sqrt[k+1]{\left(1-\dfrac{x}{k}\right)^k}$$
Simplify and find your result.
A: It's true if $k \geqslant x$ and it follows from Bernoulli's inequality
$$\left(1 - \frac{x}{k+1} \right)^{(k+1)/k} > 1 - \frac{k+1}{k}\frac{x}{k+1} = 1 - \frac{x}{k} \\ \implies \left(1 - \frac{x}{k+1} \right)^{k+1} > \left(1 - \frac{x}{k} \right)^{k}$$
A: Hint: 
$\frac{d}{dk}\left( e^{k\ln\left(1-\frac{x}{k}\right)} \right)$ is positive for all positive $k \ge 1$ [$k \in \mathbb{R}; k-1 >0; k > x$]
A: Consider $f(x) = k\ln(k-x)- k\ln k - (k+1)\ln(k+1-x) + (k+1)\ln(k+1), k>x > 0, k \ge 1\implies f'(x) = -\dfrac{k}{k-x}+\dfrac{k+1}{k+1-x}= - \dfrac{x}{(k-x)(k+1-x)}<0\implies f(x) < f(0)=0$ and this implies the result. For if $x = k$, the result is clearly true, and lastly if $x > k$ and if $k$ is odd, the $LHS < RHS$ and it is true again. If $k$ is even, can you continue to finish it off?
A: Want
$\left(1-\frac{x}{k}\right)^k
\lt\left(1-\frac{x}{k+1}\right)^{k+1}
$.
We must have 
$0 < x < k$.
Let
$f(k)
=k\ln(1-x/k)
$.
$\begin{array}\\
f'(k)
&=\ln(1-x/k)+k\dfrac{(1-x/k)'}{1-x/k}\\
&=\ln(1-x/k)+k\dfrac{x/k^2}{1-x/k}\\
&=\ln(1-x/k)+\dfrac{x/k}{1-x/k}\\
&=\ln(1-z)+\dfrac{z}{1-z}
\qquad z = x/k
\text{ so } 0 < z < 1\\
&=-\sum_{n=1}^{\infty}\dfrac{z^n}{n}+\sum_{n=1}^{\infty}z^n\\
&=\sum_{n=1}^{\infty}z^n(1-\frac1{n})\\
&=\sum_{n=2}^{\infty}z^n(1-\frac1{n})\\
&> 0\\
\end{array}
$
Therefore
$f(k)$
is an increasing function of $k$.
