first power are zero. We write: $$y(x+dx) = (x + dx)^{n} =\overbrace{(x + dx)(x + dx) \cdots
(x + dx)}^{\mbox{n\, times}} \label{eq:binom}$$ The highest power of $x$ in Eq. \[eq:binom\] comes from multiplying all of the $x's$ together. There is only one way to do this, so there is a leading term $x^{n}$.

The next term will come from multiplying together $(n-1)$ $x's$ and one $dx$. How many such terms are there? That’s an easy question, because each term omits one $dx$ and there are exactly $n$ different $dx's$ in the product in Eq. \[eq:binom\]. The second term is therefore $nx^{n-1}dx$.

What about the next term? It will have the product of $(n-2)$ $x's$, two $dx's$ and a numerical coefficient, let’s call it $a$. There will be more terms, involving products of three and even more $dx's$ (if $n$ is bigger than 2). The numerical coefficient that goes with each term is known from the [*binomial theorem*]{} that people learn about in pre-calculus courses.

But we don’t care about the value of $a$ and all those numerical coefficients because they all involve products of two or more $dx's$, and since $dx$ is a dibbl the product of two or more 