So basically, the wording in this question, to me, is weird. It goes as follows:
Explain why the following formula gives the power $e$ of a given prime $p$ in $n!$:
$$e = \sum\limits_{i=1}^{\log_pn}\left\lfloor\frac{n}{p^i}\right\rfloor.$$
Take $15!$: this function says $15! = 2^{11}\cdot 3^6\cdot 5^3\cdot 7^2\cdot 13$. This is really interesting, but I have no idea how this little function works. So if someone could give some intuition on that, it'd be much obliged!
In certain sense, this is a generic way of transforming a sum to something else which is hopefully more manageable. To understand what it is doing, let us look at following figure and count the number of black squares.
$$\begin{array}{cccc|cc} \blacksquare & \square & \square & \square & \rightarrow & 1 \\ \blacksquare & \square & \blacksquare & \square & \rightarrow & 2 \\ \blacksquare & \square & \blacksquare & \blacksquare & \rightarrow & 3 \\ \blacksquare & \blacksquare & \blacksquare & \blacksquare & \rightarrow & 4\\ \hline \downarrow & \downarrow & \downarrow & \downarrow & & \downarrow \\ 4 & 1 & 3 & 2 & \rightarrow & 10 \end{array}$$ One obvious way of counting the black squares is count the number in each column and then sum the result. This gives us $$4 + 1 + 3 + 2 \to 10.$$
Another way is slice the figure horizontally in four layers. Count the number of black squares in each layer and then sum the result. This give us $$1 + 2 + 3 + 4 \to 10\;\text{ again!}$$ The key is no matter which way you count, you get the same number.
Now imagine you pick a prime $p$ and replace the figure above by one with $n$ column and at most $\lfloor \log_p n\rfloor$ layers.
For each integer $k$ between $1$ and $n$, you place $e_k$ black squares into the bottom of $k^{th}$ column where $e_k$ is the exponent of $p$ in the prime factorization of $k$. The total number of black squares will be the number $e$ you are looking for. i.e.
$$e = \sum_{k=1}^n e_k$$
Now slice your figure horizontally and count from bottom to top.
On the $1^{st}$ layer (i.e the bottom most layer), the number of black square is the number of integer $k$ which divides $p$. There are $\lfloor \frac{n}{p^1} \rfloor$ of them.
On the $2^{nd}$ layer, the number of black sqaure is the number of integer $k$ which divides $p^2$. There are $\lfloor \frac{n}{p^2} \rfloor$ of them.
Repeat this procedure to other layers. In general, there will be $\lfloor \frac{n}{p^j} \rfloor$ black squares in the $j^{th}$ layer.
If you add this up, you will get back same number of black squares. This implies:
$$e = \sum_{k=1}^n e_k = \sum_{j=1}^{\log_p n} \left\lfloor \frac{n}{p^j} \right\rfloor$$
How many numbers in the product $1\cdot 2 \cdots \cdot n$ are divisible by $p$? How many are divisible by $p^2$? $p^3$?
Answer the above questions and it should be more clear. Start with a smaller example than $n=15$. In fact, find the smallest $n$ for which you don't understand why the formula works, and go from there.
For better intuition, let us work with particular numbers. We want the highest power of $3$ that divides $100!$, that is, $(1)(2)(3)(4)\cdots (99)(100)$.
For any number $k$ from $1$ to $100$, let $e_k$ be the highest power of $3$ that divides $k$. So for example $e_2=0$, $e_{12}=1$, $e_{63}=2$, and $e_{81}=4$. Then $$e=e_1+e_2+e_3+\cdots+e_{99}+e_{100}.$$ To view things more concretely, suppose that each number $k$ has to pay a tax of $e_k$ dollars. Then $e$ is the total tax levied by the taxation authority.
First let us gather a tax of $1$ dollar from each multiple of $3$. We gather a total of $\left\lfloor \frac{100}{3}\right\rfloor$.
Now $3$, $6$, $12$, $15$, $\dots$ have paid what they owe. But $9$, $18$, $27$, $\dots$ still owe money, Let us gather a dollar from each of them. We get $\left\lfloor \frac{100}{9}\right\rfloor$ dollars.
But $27$, $54$, $81$ still owe money. Get a dollar from each. We get $\left\lfloor \frac{100}{27}\right\rfloor$ dollars.
But $81$ is still not fully paid up. We get an additional $\left\lfloor \frac{100}{81}\right\rfloor$ dollars. So the total tax $e$ paid is $$\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor \frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor \frac{100}{81}\right\rfloor.$$
The story is the same for any prime. To see how many terms we will have in the sum, let $p^t$ be the highest power of $p$ which is $\le n$. So $t=\lfloor \log_p(n)\rfloor$. Then the highest amount of tax owed by any number in the interval $1$ to $n$ is $t$, and therefore the sum has $t$ terms.
Every $p$th integer is divisible by $p$. There are $\left\lfloor\dfrac{n}{p}\right\rfloor$ such numbers $\le n$.
Every $p^2$th integer is divisible by another factor of $p$. There are $\left\lfloor\dfrac{n}{p^2}\right\rfloor$ such numbers $\le n$.
Every $p^3$th integer is divisible by yet another factor of $p$. There are $\left\lfloor\dfrac{n}{p^3}\right\rfloor$ such numbers $\le n$.
And so on. So the total number factors of $p$ in the product of all those numbers (i.e. in $n!$) is $$\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\cdots$$
where the sum is infinite. Fortunately $\left\lfloor\dfrac{n}{p^r}\right\rfloor$ is zero for $r > \log_p n$, so we can truncate the sum there.
