Intuitively, why should the coefficient of the derivative of $x^n$ be $n$? I am able to differentiate $x^n$ with respect to $x$ from first principles using the definition of differentiation. Also it seems natural that the gradient of a finite polynomial will be one order lower. However the fact that the coefficient of the derivative of $x^5$ should be $5$, for example, seems less obvious. 

Is there a way of showing this result so that it is intuitive?
Picture a cube in $n$-space bounded by the coordinate hyperplanes and other hyperplanes parallel to them at a distance $x$ from them.  In the case $n=2$, it's easy to draw the picture: the boundaries are the two coordinate axes and two lines parallel to those, and you're looking at an $x$-by-$x$ square.

The volume of the cube is $x^n$.  As $x$ changes, how fast does the volume change?

Use what I call the "boundary rule":

\begin{align}
&amp; \text{[size of moving boundary]} \times \text{[rate of motion of boundary]} \\[6pt]
= {} &amp; \text{[rate of change of size of bounded region].}
\end{align}

There are $n$ moving boundaries, each of size $x^{n-1}$.  The rate at which each moves is the rate at which $x$ changes.  Hence the rate of change of $x^n$ is $nx^{n-1}$ times the rate of change of $x$.

PS: The boundary rule can be used to prove the product rule if you use a rectangle rather than a square.  The two moving boundaries have lengths $f$ and $g$; the rate of motion of each is the rate of change of the other.

PPS: The fundamental theorem also follows from the boundary rule (as noted in comments below).
$$
A(x)=\int_a^x f(t)\,dx.
$$
The size of the boundary is $f(x)$; the rate at which the boundary moves is the rate at which $x$ changes; hence $\dfrac{dA(x)}{dx}=f(x)$.
Take 1, and go on integrating.. $ x/1, x^2/2!, x^3/3! $..

Then take $x^n$, go on differentiating.. $ n x^{n-1}, n (n-1) x^{n-2} $ 

So there is an order with the factorial when degree is changing both in integration and 

differentiation. Intuitivly in the process $ n^ {th}$ derivative  should have $n$ at the 

head or coefficient as it is built up to that degree from unity.