Proof of $k$th derivatives always being integers Consider the function $f(x)$ defined such that $$f(x)=\frac{x^n(1-x)^n}{n!}$$
Then prove that the $k$th derivatives $f^{(k)}(0)$ and $f^{(k)}(1)$ are always integers. Here $n$ and $k$ are integers and $n\ge1$ and $k\ge 0$
I used binomial theorem to prove that the result holds true for $x=0$ for all $k\le n$ which was trivial. But I'm not able to prove it for the general case as such. I also tried using induction..but that didn't work as well.
Any research I do to find the answer online leads me to the proof of the irrationality of $\pi$ or $e^n$ where the above statement is taken as a starting point without the proof.
Thanks for any answers!! 
 A: Hint. Have you tried to differentiate this? By the general product rule we have that the $k$th derivative is $$\frac{1}{n!}\sum_{i=0}^k {k\choose j}(x^n)^{(k-j)}[(1-x)^n]^{(j)},$$ where the $i$th derivative of $y^m$ is given by $$\frac{m!}{(m-i)!}y^{m-i}.$$
A: Since $f(x)=\frac{x^n(1-x)^n}{n!}$ has the factor $x^n,$ $f^{(i)}(0)$ vanishes for $0\le i\le n-1.$ Furthermore, since $f$ is of degree $2n,$ $f^{(i)}(0)$ vanishes for $i\ge 2n+1$. Hence, we are only concerned with $f^{(i)}(0)$ for $n\le i\le 2n.$
Now, by the binomial theorem, we have $f(x)=\frac{x^n}{n!}\sum_{k=0}^n (-1)^k \frac{n!}{k!(n-k)!}x^k=\sum_{k=0}^n (-1)^k \frac{1}{k!(n-k)!}x^{k+n}.$ Note that the coefficient of $x^{k+n}$ is $\frac{f^{(n+k)}(0)}{(n+k)!},$ so we have $f^{(n+k)}(0)=(-1)^k\frac{1}{k!(n-k)!}\cdot (n+k)!,$ which you can easily prove is an integer. Hence, all the $f^{(i)}(0)$ are integers.
Now, note that since $f(x)=f(1-x),$ $f^{(i)}(0)=(-1)^if^{(i)}(1),$ hence the $f^{(i)}(1)$ are also all integers.
A: After binomial expansion, the unsigned coefficient of $x^{k}$ in the given polynomial is $\displaystyle\binom n{k-n}$. We obtain the value of the $k^{th}$ derivative at $x=0$ as the constant term generated by $x^{k}$,
$$\binom n{k-n}k!$$
which is obviously a multiple of $n!$ for $k\ge n$.

Note that this formula is valid for $k$ in $[0,n-1]$, by the zero convention on negative arguments of the binomial coefficient.
E.g., with $n=3$,
$$
1,\ \ \ \ 3,\ \ \ \ 3,\ \ \ \ 1,\ \ \ \ 0,\ \ \ \ 0,\ \ \ \ 0=0\cdot0!\\
6,\ \ \ \ 15,\ \ \ \ 12,\ \ \ \ 3,\ \ \ \ 0,\ \ \ \ 0=0\cdot1!\\
30,\ \ \ \ 60,\ \ \ \ 36,\ \ \ \ 6,\ \ \ \ 0=0\cdot2!\\
120,\ \ \ \ 180,\ \ \ \ 72,\ \ \ \ 6=1\cdot3!\\
360,\ \ \ \ 360,\ \ \ \ 72=3\cdot4!\\
720,\ \ \ \ 360=3\cdot5!\\
720=1\cdot6!
$$
A: You may try to prove the more general theorem (which appears in Ivan M. Niven, Irrational Numbers, Carus Mathematical Monograpgs, MAA, 1956, page 16):

Theorem: If $$g(x) =\frac {x^nf(x)} {n!} $$ where $f(x) $ is a polynomial with integer coefficients then $g^{(j)} (0)$ is an integer for all $j=0,1,2,\dots$. Moreover with the possible exception of $j=n$ the $g^{(j)} (0)$ is divisible by $(n+1)$. No exception is to be made if $f(0)=0$.

Since differentiation is a linear operator and a linear combination of given integers with integer coefficients is also an integer and has the divisibility by $(n+1)$ if the given integers are divisible by $(n+1)$ you can prove the above theorem by proving the case when $f(x) =x^m$, $m$ being a non-negative integer.
If $f(x) =x^m$ then $$g^{(j)} (x) =\frac{(m+n) (m+n-1)(m+n-2)\dots(m+n-j+1)}{n!}x^{m+n-j}$$ If $j<m+n$ or $j>m+n$ the derivative $g^{(j)} (0)=0$ which is obviously an integer and if $j=m+n$ then $$g^{(j)} (0)=\frac{(m+n)!}{n!}=(n+1)(n+2)\dots(n+m)$$ so that this is also an integer. And one can easily see that the derivatives are divisible by $(n+1)$ unless $m=0$ ie $f(x) =1$.

The above theorem settles your problem for derivatives at $0$. For derivatives at $1$ note that your function $f$ satisfies $f(x) =f(1-x)$ so that $f^{(j)} (0)=(-1)^jf^{(j)}(1)$.
