SARIMA model equation Can someone please tell me in the book here
 how is this SARIMA equation obtained?

I know that AR(1)=$Y_t=\alpha_1Y_{t-1}+e_t$
Non Seasonal AR(1)=>
$Y_t(1-\alpha_1B)=e_t$.  
My question is what happens to the $e_t$ in the $Y_t(1-\alpha_1B)=e_t$.   ?Can someone please show me how this equation was obtained?
 A: Excerpt from a comment clarifying the actual confusion: 

I always found it a odd way to define the model. Consider the SARIMA(1,0,0)(1,0,0)24... isn't this the same as the much simpler rendering $y_t=\beta_0+\beta_1 y_{t-1}+\beta_2 y_{t-24}+\beta_3 y_{t-25}$? Why define it in that weird way?

Now 
$$ y_t = \beta_0 + \beta_1 y_{t-1} + \beta_2 y_{t-24} + \beta_3 y_{t-25} + \varepsilon_t $$
is an ARIMA(25,0,0) model (with some coefficients set to zero). Is it the same as SARIMA(1,0,0)(1,0,0)24? Actually, no. They are the same up to a restriction on the coefficients. In SARIMA(1,0,0)(1,0,0)24, the following must hold:
$$ \beta_1 = \phi_1, \ \beta_{2} = \Phi_{24}, \ \beta_{3} = - \phi_{1}\Phi_{24}. $$ 
Hence, for a given pair $(\beta_1,\beta_{2})$, the remaining coefficient $\beta_{3}$ is fixed: $\beta_{3} = - \beta_1\beta_{2}$. If this restriction does not hold, you have an ARIMA(25,0,0) rather than SARIMA(1,0,0)(1,0,0)24. 
This is also a testable hypothesis (although I am not sure how useful such a test is from the subject-matter perspective); you may estimate an ARIMA(25,0,0) with zero restrictions on all lags but 1, 24 and 25, and test for the hypothesis $\text{H}_0: \beta_{3} = - \beta_1\beta_{2}$. If you cannot reject it, you would go for SARIMA(1,0,0)(1,0,0)24; if you reject it, ARIMA(25,0,0) would be your choice.
A: ARIMA model uses the historical information within the own variable to predict the future values and this model have two parts. On the one hand we have the AR structure and on the other hand we have the MA structure. The AR structure is very intuitive because we are multiplying one past variable's value by one estimated parameter, it's like a weight of this past in the prediction. However the MA structure it's less intuitive but it's the same intuition regarding AR. If you'd like to have a detailed explanation, tell me. 
So, pay attention to the subindex "t" in your model because I think here is the key of your question. e_t is just the error that you will get with your forecasts: Tomorrow, we'll see X sales, but when tomorrow arrives the sales have been Y (approximated value to X but different), so when you carried out the forecast you had to introduce the error term to be consistent: Y_t=X+e_t, Y is the real value at t, X the estimated one and e the difference between the real and the estimated. Y_t and e_t is the information that you have today, the X value is based on the past (based on t-1, t-2,...t-n info) and it's your work with the ARIMA model.
A: $Y(t)=α1*Y(t−1)+e(t)$
Since $Y(t-1)=B*Y(t)$ we get
$Y(t)=α1*B*Y(t)+e(t)$
Collecting like terms
$Y(t)-α1*B*Y(t)=e(t)$
or
$Y(t)[1-α1*B]=e(t)$
the forecast equation is $YPRED(t)=α1*Y(t−1)$ since the expected value of $e(t)=0.0$
