$$“(B-1)^n = B^n”$$

\begin{align*}

(B-1)^2 &= B^2\\

B^2 – 2B^1 + 1 &= B^2 \\

-2B^1+1 & = 0\\

B^1 &= \frac{1}{2} \\

B_1 &= \frac12

\end{align*}

\begin{align*}

(B-1)^3 &= B^3\\

B^3 – 3B^2 + 3B^1 – 1 &= B^3 \\

B^3 – 3B^2 + 3B_1 – 1 &= B^3 \\

-3B^2 + 3(\frac{1}{2})-1& = 0\\

-3B_2 + \frac{1}{2} &= 0 \\

B_2 &= \frac{1}{6}

\end{align*}

…

Thanks to Laurens Gunnarsen for showing me this strange trick.

I’ve finally understood the principle which allows us to lower the index. The step where we move $B^i$ to be $B_i$ is quite simple. As Rota and Roman say, one method of expressing an infinite sequence of numbers is by a transform method. That is, to define a linear transform $B$ such that $$B x^n = B_n$$

So, the above “lowering of the index” is actually using the relation $(X-1)^n = X^n$, and applying $B$ to both sides of it. To get $B(X-1)^n = B(X^n)$. Let’s look for example at the “lowering step” of the first calculation:

\begin{align*}

X^1 &= \frac{1}{2} \\

B (X^1) &= B(\frac12) \\

B_1 &= \frac12 B(1) = \frac12

\end{align*}