Is multiplication not just repeated addition? If multiplication is not repeated addition, what is it? How do you define multiplication?
All of us had to memorize the multiplication table in elementary school, but how did they come up with it if so many people claim that it is wrong to think of it as repeated addition?
Voodoo magic?
The first thing you should consider is that there are, in some sense, different "hierarchies" of numbers. At each stage, we enlarge the class of numbers, and try to do so in a way that leaves everything we could do before still the same, but that now we can do more.

A common way of doing this is to start with the natural numbers (sometimes called "counting numbers" or positive integers). We start with $1$, $2$, $3,\ldots$. 

Here, we do define multiplication as repeated addition. For example, one way to define multiplication is to assume we know how to add, and then define it by saying:
$$\begin{align*}
n\times 1 &amp;= n\\
n\times (k+1) &amp;= (n\times k) + n % need to edit at least 6 characters
\end{align*}$$
Using mathematical induction, we can show this defines multiplication for all positive integers, and that it has the usual properties we know (commutative, so that $n\times k = k\times n$ for all positive integer $n$ and $k$, distributes over the sum, is associative, etc).

Then we have two choices for "expanding our universe of numbers": we can now define negative integers, by considering things that would help us solve all equations of the form $a+x=b$ with $a$ and $b$ positive integers; or we can introduce positive rationals (fractions) by considering all things that would help us solve all equations of the form $ax = b$. Let's do the latter, since that came first historically.

So, we had the positive integers, and we knew how to add and multiply them. Now we are going to have more numbers: now, for every pair of positive integers $a$ and $b$, we will have a number "$\frac{a}{b}$", which is a number that satisfies the property that
$$b\times\left(\frac{a}{b}\right) = a.$$
We also say that $\frac{a}{b}$ is "the same fraction" as $\frac{c}{d}$ if and only if $ad=bc$ (here we are comparing products of positive integers, so that's fine).

We also notice that our old positive integers can also be considered fractions: the positive integer $a$ is a solution to $1x = a$, so $a$ corresponds to the fraction $\frac{a}{1}$.

Now, how do we add two of these numbers? Since $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sx=r$, then $\frac{a}{b}+\frac{r}{s}$ represents the solution to something; to what? A bit of algebra will tell you that it is the solution to precisely $(bs)x = (as+br)$. So we define
$$\frac{a}{b}+\frac{r}{s} = \frac{as+br}{bs}.$$
There's a bit of work that needs to be done to ensure that if you write the fractions differently, the answer comes out the same (if $\frac{c}{d}=\frac{a}{b}$, and if $\frac{t}{u}=\frac{r}{s}$, does $\frac{cu+td}{du} = \frac{as+br}{bs}$? Yes). And we also notice that if we add positive integers as if they were fractions, we get the same answer we did before:
$$\frac{a}{1} + \frac{c}{1} = \frac{a1+c1}{1} = \frac{a+c}{1}.$$
That's good; it means we are enlarging our universe, not changing it.

How about products? If $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sy=r$, their product will be the solution to $(bs)z = ar$. So we define
$$\frac{a}{b}\times\frac{r}{s} = \frac{ar}{bs}.$$
And then we notice that it extends the definition of multiplication for integers, since $\frac{a}{1}\times\frac{b}{1} = \frac{a\times b}{1}$.  And we check to see that multiplication and addition still have the properties we want (commutativity, associativity, etc).

(There are other ways to figure out what multiplication of fractions "should be", on the basis of what we want it to do. For example, we want multiplication to extend multiplication of integers, so $\frac{a}{1}\times\frac{b}{1}$ should be $\frac{ab}{1}$; and we want it to distribute over the sum, so we want
$$\frac{a}{1} = \frac{a}{1}\times \frac{1}{1} = \frac{a}{1}\times\left(\underbrace{\frac{1}{b}+\frac{1}{b}+\cdots+\frac{1}{b}}_{b\text{ summands}}\right) = \underbrace{\left(\frac{a}{1}\times\frac{1}{b}\right) + \cdots + \left(\frac{a}{1}\times\frac{1}{b}\right)}_{b\text{ summands}}.$$
So $\frac{a}{1}\times \frac{1}{b}$ should be a fraction which, when added to itself $b$ times, equals $a$; that is, a solution to $bx=a$; that is, $\frac{a}{b}$. And so on).

Then we move on from the positive rationals (fractions) to the positive reals. This is more complicated, as it involves "filling in gaps" between rationals. It is very technical. But what it turns out is that for every real number you can find a sequence of rationals $q_1,q_2,q_3,\ldots$ that get progressively closer to each other and to $r$ (we say the sequence "converges to $r$"); it won't hurt too much if you think of the $q_i$ as being progressive decimal approximations to $r$ (they don't have to be, and ahead of time you don't have any notion of decimal approximation, but you can think of it that way for our purpose). So then the way we define multiplication of real numbers $r$ and $s$ is to find a sequence of rationals $q_1,q_2,q_3,\ldots$ giving the approximation to $r$, and one $p_1,p_2,p_3,\ldots$ giving the approximation to $s$, and we define $r\times s$ to be whatever it is that the sequence
$$p_1\times q_1,\ p_2\times q_2,\ p_3\times q_3,\ \ldots$$
approximates. This ensures that if you take rational numbers and multiply them as if they were reals you get the same thing as if you multiply them as rationals, and likewise for integers.

So multiplication of positive reals is really a series of approximations made up by multiplying rationals; and multiplication of rationals is really a way to codify solutions to certain equations with integers; and it is only multiplication of (positive) integers that really corresponds to "repeated addition".

Finally, once you have the positive reals, you can introduce the negative real numbers. We consider solutions to equations of the form $a+x=b$ with $a$ and $b$ positive real numbers. Some of these already have solutions, some don't. This gives us "zero" and "negative reals". We then extend the definition of multiplication to "zero" and to "the negative reals" in a way that makes sense relative to this definition. Turns out we need to make $0\times r = 0$ for all $r$, and have to respect the "rules of signs" to make sure everything still works. So we define it that way to make sure everything works and what we had before still works exactly the same.
Devlin is profane - do not trust him!

Multiplication is repeated addition ($N = n·u= u+u+u+...$, unit is preserved) , but there is another thing called product.($u×u = u^2$, unit is changed).

Euclid had problems with this issue too, but clearly he was aware of it.
Euclid speaks of product $A\times B$ as plane number, and of product $ A \times B \times C$ as solid number.

René Descartes resolved the issue completely.

These terms is often(almost always) misused, because $|A × B| = |A| · |B|$.