Are seasonal differencing and polynomial trends interchangeable? Even though the title suggests just a simple question, it's a two-part.
First, can a repeated seasonal differencing filter also remove a polynomial trend? In which conditions? I would guess only when the degree of the trend is a multiple of the lag used in the filter. 
Second, are both the seasonal differencing and the simple (repeated) differencing interchangeable? Is working with a time series by first taking out the trend and then seasonality the same as working with a t.s. by first taking out the seasonality and then the trend?
Any help would be appreciated.
 A: I do not have general answers, but here are some thoughts that will hopefully be helpful.

First, can a repeated seasonal differencing filter also remove a polynomial trend? In which conditions? I would guess only when the degree of the trend is a multiple of the lag used in the filter.

A simple example of seasonal integration with a seasonal period $m$  is a series made up of $m$ alternating independent random walks. That is, take random walks $x_{1,t},\dots,x_{m,t}$ and form a series $y_{\tau}$ of the form
$$
x_{1,1},\dots,x_{m,1},x_{1,2},\dots,x_{m,2},x_{1,3},\dots,x_{m,3},\dots .
$$
A seasonally differenced process then will be 
$$
\Delta x_{1,2},\dots,\Delta x_{m,2},\Delta x_{1,3},\dots,\Delta x_{m,3},\dots
$$
where $\Delta x_{i,t}:=x_{i,t}-x_{i,t-1}$. So seasonal differencing applied on $y_{\tau}$ is equivalent to simple differencing applied on the alternating component series. The same is true for higher-order seasonal differencing of $y_{\tau}$ vs. higher-order simple differencing of the component series. So from this point on we can consider simple differencing of the component series instead of seasonal differencing of the original series.
If $x_{i,t}$ has a linear trend, $\Delta x_{i,t}$ will no longer have it.
If $x_{i,t}$ has a quadratic trend, $\Delta x_{i,t}$ will have a linear one, but $\Delta^2 x_{i,t}$ will have none.
And so on for higher-order polynomial trends vs. higher-order differencing.

Second, are both the seasonal differencing and the simple (repeated) differencing interchangeable? 

No, they are not intechangeable.
Consider the setup presented above.
Seasonal differencing only involves simple differencing of the alternating component series. Therefore, $x_{i,s}$ never gets "mixed up" (i.e. added to or subtracted from) $x_{j,t}$, where $i\neq j$. The different component series remain separated, and a seasonal difference of $y_{\tau}$ of order $D$ can be written as an alternating sequence of simple differences of order $D$ of the component series.
Meanwhile, simple differencing of $y_{\tau}$ "mixes up" the component series: you get values like $x_{2,1}-x_{1,1},\dots,x_{m,1}-x_{m-1,1},x_{1,2}-x_{m,1},x_{2,2}-x_{1,2},\dots,x_{m,2}-x_{m-1,2},x_{1,3}-x_{m,2},x_{2,3}-x_{1,3},\dots,x_{m,3}-x_{m-1,3}$ etc. I do not think you could "unmix" the components by taking higher order simple differences, it would rather go the other way around -- you would be involving even more different series (up to a point where all series are involved, of course).

Is working with a time series by first taking out the trend and then seasonality the same as working with a t.s. by first taking out the seasonality and then the trend?

I guess it depends on how exactly you are doing this. Algebraically this is certainly possible; you can always write $(x_t+s_t)+t=(x_t+t)+s_t$ where $s_t$ stands for the seasonal component and $t$ for a linear time trend. But when you try to achieve this using a particular model being estimated on some data, the equivalence (or lack thereof) between the fitted values of the seasonal and the trend components will depend on the model.
A: Seasonal differencing is applied once to remove a cyclical component.  This would not remove a polynomial trend such as a linear or a quadratic trend. First differencing is used to remove a linear trend.
