He's making showing that the m'th derivative of p(x) = x^n (a-bx)^n / n! is an integer at 0 and pi for 0 <= m <= 2m way harder than it needs to be.
He notes that p(x) is a polynomial of degree 2n, and then applies that general formula for multiple derivatives of a product that he proves at the start of the video. That involves a bunch of fiddling with binomial coefficients. He does this so he can write down an exact equation for the polynomial and its derivatives as a sum of powers of x. Around 20% of the video is spend on proving that general derivative formula and later applying it.
That's overkill because we do not need to know the exact coefficient of each power of x.
Simpler is to note that since p(x) is a polynomial of degree 2n and the whole thing is divided by n! it is a sum of terms of the form Ci x^i / n! for i = 0...2n. Also note that because it has a factor of x^n, Ci = 0 for i < n.
Consider then some arbitrary term of the polynomial, Ci x^i / n!, i >= n.
When x = 0, that term is 0 because of x^i.
Differentiating lowers the exponent of x by 1, so the first i-1 derivates all still have a factor of x in them, so are all 0 at x = 0.
The i'th derivative finally gets rid of the x. In particular, the i'th derivative of x^i is i!, so the i'th derivative of the term is Ci i! / n!. Remember that i >= n, so i!/n! is an integer and so is Ci i! / n!.
Further derivatives of the term are all 0 because they are derivatives of a constant.
In summary, the polynomial is 0 at x = 0, because all terms have a factor of x in them. As you repeatedly differentiate the polynomial each individual term of the derivative is 0 at x = 0 until you've differentiated enough times to get rid of all its factors of x. At that point the term is a constant integer. Further derivatives of the term are derivatives of a constant which is 0.
To show that the polynomial and its derivatives are integers at pi note that f(x) = f(pi-x), and so f'(x) = - f'(pi-x), f''(x) = f''(pi-x), ..., f^(m)(x) = (-1)^m f^(m)(pi-x). The polynomial and all its derivatives at pi are either the same as they are at 0 or the negative of what they are at 0, and since they are all integers at 0 they are also all integers at pi.
He notes that p(x) is a polynomial of degree 2n, and then applies that general formula for multiple derivatives of a product that he proves at the start of the video. That involves a bunch of fiddling with binomial coefficients. He does this so he can write down an exact equation for the polynomial and its derivatives as a sum of powers of x. Around 20% of the video is spend on proving that general derivative formula and later applying it.
That's overkill because we do not need to know the exact coefficient of each power of x.
Simpler is to note that since p(x) is a polynomial of degree 2n and the whole thing is divided by n! it is a sum of terms of the form Ci x^i / n! for i = 0...2n. Also note that because it has a factor of x^n, Ci = 0 for i < n.
Consider then some arbitrary term of the polynomial, Ci x^i / n!, i >= n.
When x = 0, that term is 0 because of x^i.
Differentiating lowers the exponent of x by 1, so the first i-1 derivates all still have a factor of x in them, so are all 0 at x = 0.
The i'th derivative finally gets rid of the x. In particular, the i'th derivative of x^i is i!, so the i'th derivative of the term is Ci i! / n!. Remember that i >= n, so i!/n! is an integer and so is Ci i! / n!.
Further derivatives of the term are all 0 because they are derivatives of a constant.
In summary, the polynomial is 0 at x = 0, because all terms have a factor of x in them. As you repeatedly differentiate the polynomial each individual term of the derivative is 0 at x = 0 until you've differentiated enough times to get rid of all its factors of x. At that point the term is a constant integer. Further derivatives of the term are derivatives of a constant which is 0.
To show that the polynomial and its derivatives are integers at pi note that f(x) = f(pi-x), and so f'(x) = - f'(pi-x), f''(x) = f''(pi-x), ..., f^(m)(x) = (-1)^m f^(m)(pi-x). The polynomial and all its derivatives at pi are either the same as they are at 0 or the negative of what they are at 0, and since they are all integers at 0 they are also all integers at pi.