Proving the inequality $\frac{n^k}{k^k}\leq\binom{n}{k}\leq\frac{n^k}{k!}$ involving binomial coefficient. What is the best way of proving inequalities including binomial coefficients in general? Lets look at this example:

$$\frac{n^k}{k^k}\leq\binom{n}{k}\leq\frac{n^k}{k!}$$

What I started with is:
$$\binom{n}{k}\frac{1}{n^k}\leq\frac{1}{k!}$$
$$\frac{n!}{k!(n-k)!}\frac{1}{n^k}\leq\frac{1}{k!}$$
$$\frac{(n-1)!}{(n-k)!n^{k-1}}\leq1$$
This is where I got stuck. How can I prove that is is lower than or equal to 1? Could you give me a hint?
 A: For $0 < k < n$,
\begin{align}
\binom{n}{k} 
&= \frac{n\times (n-1)\times (n-2)\times \cdots \times(n-k+1)}{k!}\\
&\leq \frac{n^k}{k!}\quad \text{when we replace each }n-i~\text{in the numerator with
the larger number }n\\
& \\
\binom{n}{k} 
&= \frac{n\times (n-1)\times (n-2)\times \cdots \times(n-k+1)}{k!}\\
&= \frac{n}{k}\times \frac{n-1}{k-1}\times \frac{n-2}{k-2}\times
\cdots \frac{n-k+1}{1}\\
&\geq \frac{n^k}{k^k}\quad\text{when we replace each }\frac{n-i}{k-i}~
\text{by the smaller number }\frac{n}{k}
\end{align}
To answer a question asked by the OP in the comments below,
$$\frac{n}{k}-\frac{n-i}{k-i} = \frac{nk-ni-nk + ki}{k(k-i)}
= -\frac{(n-k)i}{k(k-i)} < 0$$
since $n-k$ and $k(k-i)$ both are positive integers, and so
$\displaystyle \frac nk < \frac{n-i}{k-i}$ as claimed.
A: Just remember what a factorial is and do some cancelling:
$$\frac{(n-1)!}{(n-k)!n^{k-1}}=\frac {1\cdot2\cdot3\cdot...(n-k)(n-k+1)...(n-1)} {1\cdot2\cdot3\cdot...(n-k)n^{k-1}}=\frac {(n-k+1)...(n-1)} {n^{k-1}}$$
How many factors are in the numerator? How many in the denominator? How do those factors compare to eachother?
A: Try this for that bound:
\begin{align}
 \binom{n}{k} &= \frac{n(n-1)\dots(n-k+1)}{k!} \\
&= 1\left(1-\frac{1}n \right)\cdots\left(1-\frac{k-1}n \right) \frac{n^k}{k!}\\
&< \frac{n^k}{k!} \qquad \text{as all factors on the left are }\le 1.
\end{align}
