Inequality involving kth root of n choose k I am trying to prove the following inequality:
$\sqrt[k]{n\choose k} \le n-k+1$.
I have not managed to find anything whatsoever on this, although I am  very positive it is true.
Could there be a way to prove this involving ${n\choose k}\le(\frac{en}{k})^k$ by showing $(\frac{en}{k})^k\le (n-k+1)^k$ and taking the root?
For context: I am trying to show that for
$(1+\frac{\epsilon}{x})^n<(1+\epsilon)$, $x=\frac{n(n+1)}{2}$ would suffice.
 A: Suppose we have $n \geq k \geq i+1$ so $ k-i-1 \geq 0$ and multiply this by the first inequality. We have
\begin{eqnarray*}
k (k-i-1) & \leq & n (k-i-1) \\
n -i & \leq & nk -k^2 +k -in +ik -i = (n-k+1)(k-i) \\
\frac{n-i}{k-i} & \leq & n-k+1
\end{eqnarray*}
Now multiply this inequality together for $i=0,1, \cdots k-1$ and we have
\begin{eqnarray*}
\binom{n}{k}=\frac{n}{k}\frac{n-1}{k-1} \cdots \frac{n-i}{k-i} \cdots \frac{n-k+1}{1} \leq (n\color{red}{-}k+1)^k.
\end{eqnarray*}
Now take the $k^{th}$ root and we have the result.
A: $\sqrt[k]{n\choose k} \le n-k+1$
is equivalent to
${n\choose k} 
\le (n-k+1)^k
$.
$\begin{array}\\
{n\choose k} 
&=\dfrac{n!}{k!(n-k)!}\\
&=\dfrac{\prod_{i=0}^{k-1}(n-i)}{ k!}\\
&=\dfrac{\prod_{i=0}^{k-1}(n-i)}{\prod_{i=1}^k i}\\
&=\dfrac{\prod_{i=0}^{k-1}(n-i)}{\prod_{i=0}^{k-1} (k-i)}\\
&=\prod_{i=0}^{k-1} \dfrac{n-i}{k-i}\\
\end{array}
$
so if
$\dfrac{n-i}{k-i}
\le n-k+1
$
we are done.
This is the same as
$\begin{array}\\
n-i
&\le (k-i)(n-k+1)\\
&=k(n-k+1)-i(n-k+1)\\
\end{array}
$
or
$\begin{array}\\
i(n-k)
&\le k(n-k+1)-n\\
&= k(n-k)+k-n\\
&= (k-1)(n-k)\\
\end{array}
$
or
$i 
\le k-1
$
which is true.
A: (This is quite similar to and inspired by Donald Splutterwit's answer.)
Using
$$
1 < x \le y \quad \Longrightarrow \quad 
 \frac{y}{x} \le \frac{y-1}{x-1}
$$
it follows that for $1 \le k \le n$
$$
 \frac{n}{k} \le \frac{n-1}{k-1} \le \frac{n-2}{k-2} \le \dots 
 \le \frac{n-(k-1)}{k-(k-1)} = n-k+1
$$
and therefore
$$
\binom{n}{k} = \frac{n}{k} \cdot \frac{n-1}{k-1} \cdot \frac{n-2}{k-2}  \cdots 
  \frac{n-(k-1)}{k-(k-1)} \le (n-k+1)^k \, .
$$
and this holds for $k=0$ as well.

Your idea to show that
$$
{n\choose k}\le(\frac{en}{k})^k \le (n-k+1)^k
$$
cannot work because the second inequality does not hold for $k = n$.
