Can you help me make sense of this notation? I am reading through my calculus textbook, and came across an algebra technique that I can't decipher. The author sets up:
$$e^x  \sin(x) = \left(1 + \frac x{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)\left(x -\frac{x^3}{3!} + \cdots\right)$$
…which is clear enough. However, the next lines have me baffled.

We multiply these expressions, collecting like terms just as for polynomials:
$$\begin{align}&1+x+\tfrac12x^2+\tfrac16x^3+\cdots\\\times\quad&\;\,\phantom{1+}x\phantom{1\tfrac12x^2}\;\;-\tfrac16x^3+\cdots\\\hline&\phantom{1+}\;x+\phantom{\tfrac12}x^2+\tfrac12x^3+\tfrac16x^4+\cdots\\+\quad&\phantom{1+x+\tfrac12x^2}-\tfrac16x^3-\tfrac16x^4-\cdots\\\hline&\phantom{1+}\;x+\phantom{\tfrac12}x^2+\tfrac13x^3+\cdots\end{align}\\e^x\sin x=x+x^2+\tfrac13x^3+\cdots$$

I see that he multiplied through by $x$. I do not see where $-\frac16x^4$ comes from, nor why he changes to addition. I was never taught this method of multiplication.
 A: The common multiplication algorithm for integers looks something like this:
$$\begin{align}1234&5\\\times678&9\\\hline9(1234&5)\\+80(1234&5)\\+700(1234&5)\\+6000(1234&5)\end{align}$$
What the author did there is equivalent. You have
$$x(1+x+\frac{x^2}{2}+\cdots)-\frac{x^3}{6}(1+x+\frac{x^2}{2}+\cdots)+\cdots$$
A: As per Wojowu this is just long multiplication.  But it is confusing because he omitted the brackets from the first part (where the multiplication is).  Also he ends up aligning the same power terms vertically for the adding part.
So in other words... take the first term from the second set of terms (x) and multiply it through on all the terms in the first set of terms, as per long multiplication: x * 1 = x; x * x/1! = x * x = x^2; x * (x^2)/2! = x * 1/2 * x^2 = 1/2 * x^3, etc...
Then take the second term from the second set of terms (-(x^3)/3!) and multiple it through on all the terms in the first set of terms: -(x^3)/3! * 1 = -(x^3)/3!; -(x^3)/3! * x/1! = -(x^4)/6, etc...
And so on, and line up same power terms vertically to assist with adding all these resulting terms together...
So it is just long multiplication after all, just laid out a bit differently.  Does that help?
A: During the expansion, you will get this:
$$(\frac{x}{1!})(-\frac{x^3}{3!}) $$
$$(\frac{x}{1})(-\frac{x^3}{3×2×1}) $$
$$-\frac{x^4}{6}$$
A: We have $$e^{x} \sin(x)=\sum_{n=0}^{+\infty} \frac{x^n}{n!} \times \sum_{m=0}^{+\infty} \frac{(-1)^m x^{2m+1}}{(2m+1)!}$$
$$=\sum_{n=0}^{+\infty}c_{n}x^{n},$$ where $$c_n=\sum_{k=0}^{n} \frac{(-1)^k}{(2k+1)!(n-k)!}.$$ So, for $n=4$ we get $c_4=\sum_{k=0}^{4} \frac{(-1)^k}{(2k+1)!(4-k)!}=-1/6.$
