Determinant of matrix with binomial coefficients entries

This problem is the second of three that appear in the section "Can you answer this? of Mathematics Stack Exchange Weekly Newsletter - Tuesday, November 14, 2017. In truth the biggest difficulty of this problem (at least for me) lies in being able to visualize clearly the type of matrix at stake. I start my answer with a clarifying example for those who have not seen clear as I did at the beginning. We note $A_n$ a matrix $n\times n$.

Example.-Prove that the determinant of the matrix (note that always $\binom mm=1$).

$A_5=\left( \begin{array}{ccccc} \binom{k+1}{k}x & 1 & 0 &0&0 \\\\ \binom{k+2}{k}x^2 & \binom{k+2}{k+1}x & 1&0&0\\\\\binom{k+3}{k}x^3& \binom{k+3}{k+1}x^2 & \binom{k+3}{k+2}x&1&0 \\\\ \ \binom{k+4}{k}x^4 & \binom{k+4}{k+1}x^3 &\binom{k+4}{k+2} x^2 &\binom{k+4}{k+3}x &1 \\\\\binom{k+5}{k}x^5&\binom{k+5}{k+1}x^4&\binom{k+5}{k+2}x^3&\binom{k+5}{k+3}x^2& \binom{k+5}{k+4}x \end{array}\right)$

Is equal to $\binom{k+5}{k}x^5$.

A matrix of the form $A_5=\left( \begin{array}{ccccc} a_{11}x & 1 & 0 &0 & 0 \\\\ a_{21}x^2 & a_{22}x & 1&0&0\\\\a_{31}x^3 &a_{32}x^2 &a_{33}x&1 & 0 \\\\ \ a_{41}x^4 & a_{42}x^3 & a_{43}x^2 &a_{44}x & 1 \\\\ a_{51}x^5& a_{52}x^4 & a_{53}x^3 &a_{54}x^2&a_{55}x \\ \end{array}\right)$

lends itself to apply the second principle of induction in the calculation of its determinant which is equal to

$a_{55}x\cdot \det\left( \begin{array}{ccccc} a_{11}x & 1 & 0 &0 \\\\ a_{21}x^2 & a_{22}x & 1&0\\\\a_{31}x^3 &a_{32}x^2 &a_{33}x&1 \\\\ \ a_{41}x^4 & a_{42}x^3 & a_{43}x^2 &a_{44}x \\ \end{array}\right)-\det\left( \begin{array}{ccccc} a_{11}x & 1 & 0 &0 \\\\ a_{21}x^2 & a_{22}x & 1&0\\\\a_{31}x^3 &a_{32}x^2 &a_{33}x&1 \\\\ a_{51}x^5& a_{52}x^4 & a_{53}x^3 &a_{54}x^2 \\ \end{array}\right)$

so we get iterating $$\det(A_5)=a_{55}x\det(A_4)-a_{54}x^2\det(A_3)+a_{53}x^3\det(A_2)-(a_{11}a_{52}-a_{51})x^5$$ which for the matrix $A_5$ with binomial coefficients give $\det(A_5)$ equal to

$\left(\binom{k+5}{k+4}\binom{k+4}{k}-\binom{k+5}{k+3}\binom{k+3}{k}+\binom{k+5}{k+2}\binom{k+2}{k}-\binom{k+5}{k+1}\binom{k+1}{k}+\binom{k+5}{k}\right)x^5$.

It is easy to verify that the first four terms have sum zero. In fact $\binom{k+5}{k+4}\binom{k+4}{k}-\binom{k+5}{k+3}\binom{k+3}{k}+\binom{k+5}{k+2}\binom{k+2}{k}-\binom{k+5}{k+1}\binom{k+1}{k}=0$ because it is equal to $$(k+1)(k+2)(k+3)(k+4)(k+5)\left(\frac{1}{4!}-\frac{1}{2!\cdot3!}+\frac{1}{3!\cdot2!}-\frac{1}{4!}\right)=0$$ Thus (we have applied the second principle of induction up to $A_5$) $$\det(A_5)=\binom{k+5}{k}x^5$$

$$*****$$ So we must have for $A_n$

$\frac{\det(A_n)}{x^n}=\left[\binom{k+n}{k+n-1}\binom{k+n-1}{k}-\binom{k+n}{k+n-2}\binom{k+n-2}{k}+\cdots \mp\binom{k+n}{k+2}\binom{k+2}{k}\pm\binom{k+n}{k+1}\binom{k+1}{k}\right] \pm\binom{k+n}{k}$.

When $n$ is odd (even) the last term is positive (negative).

Each of the terms $\dbinom{k+n}{k+n-r}\dbinom{k+n-r}{k}$ is equal to $\dfrac{M_n }{r!(n-r)!}$ where $M_n=(k+1)(k+2)\cdots(k+n)$. Calculation gives now $$\frac{\det(A_n)}{x^n}=M_n\left(\frac{1}{1!(n-1)!}-\frac{1}{2!(n-2)!}+\cdots+\frac{1}{(n-2)!1!} -\frac{1}{1!(n-1)!}+\frac{1}{n!}\right);\text{ n odd}$$ $$\frac{\det(A_n)}{x^n}=M_n\left(\frac{1}{1!(n-1)!}-\frac{1}{2!(n-2)!}+\cdots-\frac{1}{(n-2)!1!}+\frac{1}{1!(n-1)!}-\frac{1}{n!}\right);\text{ n even}$$ ►For $n$ odd the first $(n-1)$ summands enclosed in square brackets above give an evident zero sum so we have $$\color{red}{\det(A_n)=\frac{M_nx^n}{n!}=\binom{k+n}{k}}$$ ►For even $n=2m$ the sum enclosed in parenthesis we need to be equal to $\dfrac{1}{(2m)!}$ instead of $-\dfrac{1}{(2m)!}$ so we must have the first $(2m-1)$ terms be equal to $\dfrac{2}{(2m)!}$. Proving this $$\frac{1}{1!(2m-1)!}-\frac{1}{2!(2m-2)!}\cdots+\frac{1}{m!m!}\cdots+\frac{1}{(2m-1)!1!}=\frac{2}{(2m)!}$$ It is equivalent to $$\binom{2m}{1}-\binom{2m}{2}-\cdots+\binom{2m}{m}-\cdots+\binom{2m}{2m-1}=2\tag{*}$$ which can be deduced from $(1-1)^{2m}=0$; in fact $$(1-1)^{2m}=1-(LHS \text{ of } (*))+1=0\Rightarrow LHS \text{ of } (*)=2$$The proof is finished.