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?
 A: If you think about $x^n$ as the volume, in $n$ dimensions, of a cube of side $n$, you can ask "how does that grow when $x$ increases?" Answer: count the number of sides of dimension $n-1$. For a square in the plane, with one corner fixed at the origin, you have the change in area being produced by the upper and right edges, each multiplied by a thickness $\Delta x$, when you change $x$ to $x + \Delta x$. For a cube in 3-space, you have three sides, each of whose areas is multiplied by $\Delta x$ to get the change in volume. For a segment in $\mathbb R^1$, you have the right hand point moving through a distance $\Delta x$ to get the change in length, and so on. 
In general, there are $n$ "sides" of a hypercube in dimension $n$ (with one corner fixed at the origin), corresponding to incrementing each coordinate individually. (This is also where the $n$ in the binomial expansion of $(x + \Delta x)^n$ comes from, of course.) 
A: 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}
& \text{[size of moving boundary]} \times \text{[rate of motion of boundary]} \\[6pt]
= {} & \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 $\dfrac{dA(x)}{dx}=f(x)$.
A: Arguing with infinitesimals and using Newton's binomial: $$\frac {1}{h}((x+h)^n-x^n)=\frac{1}{h}(x^n+nhx^{n-1}+h^2(\ldots )-x^n)= nx^{n-1} + h(\dots ).$$
A: I will try to give intuition about the product rule and apply it to your case. Suppose you are taking the derivative of $f_1 \cdot f_2 \cdot \ldots f_n$ at $x$. This is roughly $\frac{1}{\Delta}$ times
$$f_1(x+\Delta) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x).$$
Now we break up this expression into a telescoping sum:
$$\big( f_1(x+\Delta) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta)  - f_1(x) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) \big) + \big( f_1(x) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x+\Delta) \big) + \ldots + \big( f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x+\Delta)  - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x) \big),$$
with each term in the outer sum successively replacing one $x+\Delta$ with $x$. This step introduces $n$ outer terms, which is where the factor of $n$ will come from. Simplifying, we get:
$$ (f_1(x+\Delta) - f_1(x)) f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) + f_1(x) (f_2(x+\Delta) - f_2(x)) \cdot \ldots \cdot f_n(x+\Delta) + \ldots + f_1(x) f_2(x) \cdot \ldots \cdot (f_n(x+\Delta) - f_n(x)),$$
which is approximately equal to
$$ (f_1(x+\Delta) - f_1(x)) f_2(x) \cdot \ldots \cdot f_n(x) + f_1(x) (f_2(x+\Delta) - f_2(x)) \cdot \ldots \cdot f_n(x) + \ldots + f_1(x) f_2(x) \cdot \ldots \cdot (f_n(x+\Delta) - f_n(x)).$$
When all the $f_i$s are equal this is simply $n (f(x+\Delta) - f(x)) f^{n-1}(x)$.
A: Solomonoff's Secret nailed the intuition by suggesting the product rule!
$$\frac{d}{dx}x^n = \frac{d}{dx}[\underbrace{(x)(x) ... (x)}_\text{n times}] \\ = \frac{dx}{dx}{\underbrace{(x)(x) ... (x)}_\text{n-1 times}} + (x)\frac{dx}{dx}\underbrace{(x)(x) ... (x)}_\text{n-2 times} + (x)(x)\frac{dx}{dx}\underbrace{(x)(x) ... (x)}_\text{n-3 times} + ... \\ = (1){\underbrace{(x)(x) ... (x)}_\text{n-1 times}} + (x)(1)\underbrace{(x)(x) ... (x)}_\text{n-2 times} + (x)(x)(1)\underbrace{(x)(x) ... (x)}_\text{n-3 times} + ... \\ =\underbrace{x^{n-1} + x^{n-1} + x^{n-1} + ... + x^{n-1}}_\text{n times} \\ = nx^{n-1}.$$
A: It comes from the very definition of the derivative in this case and the remarkable identity for $x'^n-x^n$. Indeed  the  derivative is the limit, as $x'\to x$ of
$$\frac{x'^n-x^n}{x'-x}=\sum_{k=0}^{n-1}x'^{n-k}x^k.$$
Each term in this sum tends to $x^{n-1-k}x^{k}=x^{n-1}$, and there are $n-1$ terms.
A: If
$f(x) = x^n$,
$\ln f(x) = n \ln x$.
Since
$(\ln f(x))'
=\frac{f'(x)}{f(x)}
$
and
$(\ln f(x))'
=(n \ln x)'
=\frac{n}{x}
$,
we get
$f'(x)
=f(x)(\ln f(x))'
= x^n \frac{n}{x}
=n x^{n-1}
$.
A: 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.
