Why is the $n$'th derivative of $(x-a_1)(x-a_2)\cdots (x-a_n)$ equal to $n!$?

Question

$a_1,a_2,\ldots, a_n$ are any real numbers.

Using the product and chain rules doesn't seem to work well. Thanks.

Answer 1

Let : $$p(x)= (x-a_1)(x-a_2)(x-a_3) \cdots (x-a_n)$$

On multiplying all the brackets, you'll get :

$$p(x) = x^n-b_0x^{n-1} + \dots + b_{n-2}x+ b_{n-1} $$

Where $b_i$s are some constants. $$p^1(x) = n(x)^{n-1} -(n-1) \cdot b_0x^{n-2} + \dots + b_{n-2}+0 $$

So on..

$$p^n(x)= n(n-1)(n-2) \cdots 3\cdot 2\cdot 1 \cdot x^0 = n!$$

Here $p^i(x)$ denotes $i^\text{th}$ derivative of $p(x)$.