How much do you know about the entries in Pascal's Triangle? Do you know their relation to the coefficients of the binomial expansion?
The $n+1$st row of Pascal's Triangle gives you the coefficients of the binomial $(a+b)^n$. That is, if the row is $k_0\quad k_1\quad k_2\quad\cdots\quad k_n$ that means that if you multiply out $(a+b)^n$, you will get $k_0a^n + k_1a^{n-1}b + k_2a^{n-2}b^2 + \cdots + k_nb^n.$
The coefficients are given by the binomials: $k_i = \binom{n}{i}$, where $\binom{n}{i} = \frac{n!}{i!(n-i)!}$ with $!$ symbolising the factorial, $r!=1\times 2\times\cdots \times r$. It is not hard to work out that $\binom{n}{m} = \frac {n(n-1)(n-2)\cdots(n-m+1)}{m(m-1)(m-2)\cdots 1}.$ That is, in the numerator you multiply the integers from $n$ down to $n-m+1$ ($m$ factors), and in the denominator you multiply the integers from $m$ down to $1$ (also $m$ factors).
So, suppose you already know the value $v_{c-1}$ in row $R$. This is $\binom{R-1}{c-1} = \frac{(R-1)(R-2)\cdots(R-c)}{(c-1)(c-2)\cdots 1}.$ The next value, $v_{c}$, is $\binom{R-1}{c}$, which is: \begin{align*} v_c&=\binom{R-1}{c}\\ & = \frac{(R-1)(R-2)\cdots(R-c-1)}{c(c-1)(c-2)\cdots 1}\\ &= \frac{(R-1-c)\Bigl((R-1)(R-2)\cdots(R-c)\Bigr)}{c\Bigl((c-1)(c-2)\cdots 1\Bigr)}\\ &= \frac{R-1-c}{i}\left(v_{c-1}\right). \end{align*} Now, remember this was in row $R$. So letting $r = \mathrm{row}-1 = R-1$, we get $v_c = \binom{r}{c} = v_{c-1}\left(\frac{r-c}{c}\right),$ exactly the formula given.
Added. In retrospect, this was very silly of me. If you knew that the $i$th (starting with $0$) entry in the $R$th row is given by $\binom{R}{i}$, then why would you need to remember this other formula?
So, why is the entry given by $\binom{R}{i}$? First: $\binom{R}{i}$ counts the number of ways in which you can choose $i$ things from $R$ possibilities. So, for example, $\binom{4}{2}$ tells you in how many ways you can choose two things out of four possible ones: if the four options are 1, 2, 3, and 4, you could pick 1 and 2; or 1 and 3; or 1 and 4; or 2 and 3; or 2 and 4; or 3 and 4; that is, there are six possible ways of picking two out of four. So $\binom{4}{2}=6$.
Why does the formula I gave above give the right count? If you have $R$ things to choose from, you have $R$ ways to pick out the first choice; then $R-1$ ways to pick out the second; then $R-2$ to pick out the third, and so on. You end up with $R(R-1)(R-2)\cdots(R-i+1)$ ways of choosing $i$ things. However, each choice was counted many times. In the example I gave above, I don't count "1 and 2" as different from "2 and 1". How many ways did I count each choice? Of the $i$ choices I made, I could have selected any of the $i$ things first; then any of the remaining $i-1$ second; then any of the remaining $i-2$ as my third choice, etc. That is, each choice of $i$ things can be made in $i(i-1)(i-2)\cdots (2)(1)$ ways. So, each choice was counted that many times; to get the right count, we divide the original count by the factor by which we overcounted, so $\binom{R}{i}=\frac{R(R-1)(R-2)\cdots(R-i+1)}{i(i-1)\cdots 2\times 1}.$ This is the formula I gave above.
One small wrinkle: what happens when $i=0$? Well, in how many ways you make no choices? Just one: don't do anything. So $\binom{R}{0}=1$ always.
And, why is this "number of ways to pick" related to the coefficients of $(a+b)^n$? Imagine that you need to do the product. Write them down one after the other, as if you were writing down a sum: \begin{align*} a &+b\\ a &+b\\ a &+b\\ &\vdots\\ a &+ b\\ a &+b \end{align*} You need to do all the possible products of each element with the elements of the other rows.
How many times will you get $a^n$? Just once, because the only way to get $a^n$ is by multiplying all the $a$s together. How many times will you get $a^{n-1}b$? Well, you get $a^{n-1}b$ if you pick one $b$ in one row, and multiply it by all the $a$s in the other rows. You have $n$ different choices for the one $b$, so there are $\binom{n}{1}$ ways of getting $a^{n-1}b$. How many ways can you get $a^{n-2}b^2$? You have to pick two which rows will provide you with the two $b$s. There are $n$ possibilities, you need to pick $2$, so it's $\binom{n}{2}$ ways.
And so on. How many times will you get $a^{n-i}b^i$? You need to pick $i$ rows to provide you with the $b$s, and there are $\binom{n}{i}$ ways to pick them.
So when we write out $(a+b)^n$, the monomial $a^{n-i}b^i$ will show up $\binom{n}{i}$ times. So that's the coefficient.
This means that the $i$th entry in the $(n+1)$st row of Pascal's triangle is $\binom{n}{i}$, which agrees with what we said above.
And now that you know this, this is really the easy formula to remember, not the "magic formula" form Wikipedia.