What's $(1-B)^d(a_0+...+a_{d}t^d)$? If $B$ is the lag operator or backshift operator used in time series, what's $(1-B)^d(a_0+...+a_{d}t^d)$? We can think of the polynomial as a trend in a time series process.
More specifically, what's $Bt^d$? $t^d-1$ or $(t-1)^d$?
 A: Because there are important differences between polynomials and time series, let's look at this abstractly.
Concepts and Notation
Consider a set $S$, a vector space $V$, and the collection of all functions $V^S = \{a:S\to V\}$.  The two examples to keep in mind are both where $S=\mathbb Z$ is the set of integers and $V$ is either (a) $\mathbb R$ or (b) the (real-valued) random variables defined on a given probability space $(\Omega, \mathfrak F, \mathbb P)$ (which I will simply write as "$\Omega$" below).  In this situation, elements $a$ of $V^S$ are often called sequences, their values at integers $i$ are written $a_i$ instead of $a(i)$, and are expressed in the form $$a = (a_i)_{i\in\mathbb Z} = (\ldots,\ a_{-2},\ a_{-1},\ a_0,\ a_1,\ a_2,\ \ldots).$$
The "obvious" rules for scalar multiplication and vector addition, namely
$$(\lambda a)_i = \lambda (a_i)$$
and
$$(a+b)_i = a_i + b_i$$
(for any $\lambda\in\mathbb R$ and $a,b,\in V^S$) make $V^S$ into a vector space.
Further suppose there is an injective map $\sigma:S\to S$, a so-called "shift."  The one to have in mind is $\sigma:\mathbb Z\to\mathbb Z$ given by $\sigma(i)=i+1$.  This induces a map on from $V^{\sigma(S)}$ to $V^S$, which (for clarity) I will write $[\sigma]$, given by $$[\sigma](a)_i = a_{\sigma^{-1}(i)}.$$  Explicitly, for our example with $S=\mathbb Z$, because $\sigma^{-1}(i)=i-1$, the sequence
$$-2\to a_{-2}, -1\to a_{-1}, 0\to a_0, 1\to a_1,  2\to a_2,\ldots$$
is mapped to the sequence $$-2\to a_{-3}, -1\to a_{-2}, 0\to a_{-1}, 1\to a_0, 2\to a_1, \ldots.$$  If you were to tabulate your sequences, like thus,
$$a: \left(\array{\ldots & -2 & -1 & 0 & 1 & 2 & \ldots \\ \ldots & a_{-2} & a_{-1} & a_0 & a_1 & a_2 & \ldots }\right),$$
then applying this "backshift operator" $[\sigma]$ appears to reach back one step and pull the sequence one step to the right, assigning the immediately preceding value to the current index:
$$[\sigma](a): \left(\array{\ldots & -2 & -1 & 0 & 1 & 2 & \ldots \\ \ldots & a_{-3} & a_{-2} & a_{-1} & a_0 & a_1 & \ldots }\right).\tag{1}$$
When $i$ is a time index, this procedure "looks back" one step in time. A key point is that $[\sigma]$ is a linear vector space map $[\sigma]:V^{\sigma(S)}\to V^S$.  This gives us a large set of tools from linear algebra and functional analysis for studying $[\sigma]$.
Applications
We may understand a polynomial
$$p = p_0 + p_1T + p_2 T^2 + \cdots + p_d T^d$$
(where "$T$" is merely an abstract symbol) as determining a function, also written $p$, from $\mathbb Z\to \mathbb R$ via
$$p_t = p_0 + p_1 t + p_2 t^2 + \cdots + p_d t^d$$
for all $t\in \mathbb{Z}$. (It is conventional to use "$t$" as an index in this context to remind us of the application where $t$ indexes regularly spaced times.) The "backshift operator" $B$ refers to the induced action of $\sigma$ on $V^S = \mathbb{R}^\mathbb{Z}$.  As such, definition $(1)$ yields
$$(B(p))_t = ([\sigma](p)_t) = p_{\sigma^{-1}(t)} = p_{t-1} = p(t-1) = p_0 + p_1(t-1) + \cdots + p_d(t-1)^d.$$
This answers your "more specific" question (and the rest is just a matter of algebra, working in the ring of linear operators).  But let's go just a little further, because now the abstractions begin to pay off: we can immediately see the connection with time series.
As another application, let $X=(X_t)$ be a time series: it's a sequence of random variables indexed by $\mathbb Z$.  Now
$$(B(X))_t = X_{t-1}$$
is seen as the usual backshift operator for time series..
Uniting these applications is the idea that by taking the expectations of the $X_t$ for each $t$ separately, we map the space of time series $\Omega^\mathbb Z$ to the space of real-valued sequences $\mathbb R^\mathbb Z$ via
$$E[(X)]_t = (E[X])_t.$$
Obviously this (linear) map commutes with the backshift operator: $$E \circ B = B \circ E.$$  There's nothing profound about this; I have simply stated that it doesn't matter whether you shift the sequences before or after taking expectations--that's the result I hope is obvious.  Note, though, the abuse of notation: the two instances of "$B$" in the preceding statement are linear operators on different spaces.  If there's any chance of confusion--or when you're working through these ideas for the first time--it helps to remind yourself of these distinctions.
A: Use induction.  This might be more long-winded than it needs to be.  I'll take for granted the linearity of the differencing operator $(1-B)$.
First, we show that 
$(1-B)^m t^n = 0$ if $m>n$ and $n \geq 0$
Obviously, if $n=0$, then $t^n=1$, and so $(1-B)1=0$, and any further differencing also gives zero: $(1-B)^m 1=(1-B)^{m-1} 0=0$.  That's the first bit.
Now assume that $(1-B)^m t^k=0$ if $m>k$ for $k=0 \ldots n-1$.  Then if $m>n$,
$(1-B)^m t^n = (1-B)^{m-1} \left( t^n - (t-1)^n \right) = (1-B)^{m-1} \left( t^n - (t^n + b_1 t^{n-1} + \ldots + b_n) \right)$
where the coefficients $b_i$ are given by the binomial theorem but we don't really care what they are exactly here.  Then
$(1-B)^m t^n = (1-B)^{m-1} \left( - b_1 t^{n-1} - \ldots - b_n) \right)$
But $m-1>n-1>n-2>\ldots$, and by assumption $(1-B)^m t^k=0$ if $m>k$ for $k=0 \ldots n-1$.  Therefore 
$(1-B)^m t^n = 0$ if $m>n$.  That's the second bit.
Next, we show that 
$(1-B)^n t^n = n!$ for $n>0$.  
again by induction.  
If $n=1$ then we have $(1-B) t = t - (t-1) = 1 = 1!$, so that's the first bit.  
Now assume that $(1-B)^{n-1} t^{n-1} = (n-1)!$.  Then
$(1-B)^n t^n = (1-B)^{n-1} \left(t^n - (t-1)^n \right) = (1-B)^{n-1} \left(t^n - (t^n + n t^{n-1} + b_2 t^{n-2} + \ldots + b_n \right)$
where here we care about the first coefficient in the binomial expansion ($b_1=n$) but not the rest.
$(1-B)^n t^n = (1-B)^{n-1} \left(n t^{n-1} - b_2 t^{n-2} - \ldots + b_n \right)$
By the lemma that we proved at the start, 
$(1-B)^n t^n = (1-B)^{n-1} \left(n t^{n-1} - b_2 t^{n-2} - \ldots + b_n \right) = (1-B)^{n-1} n t^{n-1}$
By the assumption used in the second bit of induction, then, we have
$(1-B)^n t^n = (1-B)^{n-1} n t^{n-1} = n (n-1)! = n!$
which completes that part of the proof.
What you're looking for then follows directly:
$(1−B)^d (a_0+\ldots+a_d t^d) = (1-B)^d a_d t^d$
by the first part, and
$(1−B)^d (a_0+\ldots+a_d t^d) = (1-B)^d a_d t^d = a_d d!$
by the second part.
