How to calculate the following union easily? If we have any sets $M_0, ... , M_n$ with $M_0 = \emptyset $. If I have the formula: $ \bigcup \limits _ {i=1}^n (M_i \setminus  \bigcup \limits_{j=0}^{i-1}M_j)$ that is equal to $\bigcup \limits_{i=1}^n M_i$ (thank's to Bungo). Is the following inductive proof correct? 
for $n=1 \qquad $ $\bigcup \limits_{i=1}^1 M_i$ = $M_1$ = $M_1 \setminus M_0$ = $ \bigcup \limits _ {i=1}^1 (M_i \setminus  \bigcup \limits_{j=0}^{0}M_0)$  
induction hypothesis: $\bigcup \limits_{i=1}^{n-1} M_i = \bigcup \limits _ {i=1}^{n-1} (M_i \setminus  \bigcup \limits_{j=0}^{i-1}M_j)$ 
induction step: 
$\bigcup \limits_{i=1}^{n} M_i = \bigcup \limits _ {i=1}^{n} (M_i \setminus  \bigcup \limits_{j=0}^{i-1}M_j)$ 
$=M_n \cup\bigcup \limits_{i=1}^{n-1} M_i$ 
according to induction hypothesis
$=M_n \cup \bigcup \limits _ {i=1}^{n-1} (M_i \setminus  \bigcup \limits_{j=0}^{i-1}M_j)$ 
$=\bigcup \limits _ {i=1}^n (M_i \setminus  \bigcup \limits_{j=0}^{i-1}M_j)$
Especially at the last step, is it allowed to sum up like this?
 A: As mentioned in the comments, induction does not seem like a fruitful way to prove this equality. All you have managed to do is to transform the original equality you wanted to prove into another one which you now have to prove (also by induction?)
Here is a direct proof which does not use induction. If for some reason you are required to use induction, hopefully you can use the same ideas as in this proof.
The goal is to prove that
$$\bigcup_{i=1}^{n}M_i = \bigcup_{i=1}^{n}(M_i \setminus \bigcup_{j=0}^{i-1}M_j)$$
To prove equality of two sets $S$ and $T$, we prove that $S \subseteq T$ and $T \subseteq S$.
Suppose that $x \in \bigcup_{i=1}^{n}M_i$. Then $x \in M_i$ for some $i$, where $1 \leq i \leq n$. Choose the smallest $i$ for which $x \in M_i$. Then $x \in M_i$ but $x \not\in \bigcup_{j=0}^{i-1}M_j$. Therefore, $x \in M_i \setminus \bigcup_{j=0}^{i-1}M_j$, and so $x \in \bigcup_{i=1}^{n} (M_i \setminus \bigcup_{j=0}^{i-1}M_j)$. This shows that $\bigcup_{i=1}^{n}M_i \subseteq \bigcup_{i=1}^{n} (M_i \setminus \bigcup_{j=0}^{i-1}M_j)$.
Now suppose that $x \in \bigcup_{i=1}^{n} (M_i \setminus \bigcup_{j=0}^{i-1}M_j)$. Then $x \in M_i \setminus \bigcup_{j=0}^{i-1}M_j$ for some $i$ with $1 \leq i \leq n$. Therefore, $x\in M_i$, and consequently $x \in \bigcup_{i=1}^{n}M_i$. This shows that $\bigcup_{i=1}^{n} (M_i \setminus \bigcup_{j=0}^{n-1}M_j) \subseteq \bigcup_{i=1}^{n}M_i$.
We have proved containment in both directions, so we're done.
A: The equations in the induction step (as you initially wrote it, at least)
are not quite as obvious as you seem to think.
In particular, the last equation seems a bit question-begging
(that is, it assumes some of what you were supposed to prove).
If you want to do this inductively, I suggest the following.
First, you should already know that for any sets $A$ and $B$,
$$ (A \setminus B) \cup B = A \cup B.$$
If this fact has not already been established, prove it now.
You can use the same induction hypothesis you are already using: 
$$\bigcup \limits_{i=1}^{n-1} M_i
= \bigcup_{i=1}^{n-1} \left(M_i \setminus \bigcup_{j=0}^{i-1}M_j\right),$$
but the rest of the induction step can proceed like this: 
\begin{align} \require{cancel}
\bigcup_{i=1}^{n} M_i
&= \xcancel{\color{red}{\bigcup_{i=1}^{n} \left(M_i \setminus \bigcup \limits_{j=0}^{i-1} M_j\right)}} \\
&=M_n \cup\bigcup_{i=1}^{n-1} M_i \\
&= \xcancel{\color{red}{M_n \cup \bigcup_{i=1}^{n-1} \left(M_i \setminus  \bigcup_{j=0}^{i-1} M_j\right)}}
 \qquad & \color{red}{\text{inductive hypothesis}} \\
&=\color{green}{\boxed{\left(M_n \setminus \bigcup_{i=1}^{n-1} M_i\right) \cup \bigcup_{i=1}^{n-1} M_i}} 
\qquad & (A \setminus B) \cup B = A \cup B\\
&=\color{green}{\boxed{\left(M_n \setminus \bigcup_{i=1}^{n-1} M_i\right) \cup \bigcup_{i=1}^{n-1} \left(M_i \setminus \bigcup_{j=0}^{i-1}M_j\right)}}
 \qquad & \text{inductive hypothesis} \\
&=\bigcup_{i=1}^n \left(M_i \setminus \bigcup_{j=0}^{i-1} M_j\right)
\end{align}
This is based on what you wrote, but
the parts that are shown in red and crossed off are parts of that
writeup that I would remove, and the parts shown in green (and in boxes)
are parts I would add.
In particular, on the first line I would not write the part on the right-hand side, because that is the conclusion you're trying to reach.
It's helpful to write this on a piece of scrap paper, perhaps, because
you do need to remember to get there eventually, but it's bad form to
put it where it appears to be a step of the derivation.
It makes the "proof" look as if you're assuming the thing you are
supposed to be proving.
It would be OK, but unnecessary, to insert an extra paragraph before
you prove the inductive step, saying something like, "To prove the inductive step we need to show that (insert formula here)."
But if you do that you have to then start the actual derivation of the
inductive step all over again with a new equation with 
$\bigcup_{i=1}^{n} M_i$ on the left-hand side.
The second thing I recommended to delete is actually a perfectly
fine application of the inductive hypothesis.
I deleted it not because it's wrong, but because it results in an
awkward expression.
At some point I want to apply the fact that
$(A \setminus B) \cup B = A \cup B$,
but if I apply the inductive step first then the thing I have to
use for $B$ is a much bigger, more complicated thing than I would like.
Even so, I think the proof would still be salvageable, but 
considerably more complicated than necessary.
A: Yes you can prove it by induction. I don't understand how you get from your 2nd to last to your last line, so let's fix that.
Picking up at the end of your argument, here's the induction hypothesis:
$$
\bigcup_{i=0}^{n-1} M_i = \bigcup_{i=0}^{n-1} (M_i\setminus \bigcup_{j=0}^{j-1}M_j).\tag{IH}
$$
Now,
$$\begin{align}
\bigcup_{i=0}^{n} M_i &= M_n \cup \bigcup_{i=0}^{n-1} M_i \\
&= (M_n\setminus \bigcup_{i=0}^{n-1} M_i) \cup \bigcup_{i=0}^{n-1} M_i \quad\textit{because for all A,B: $(A\setminus B)\cup B = A\cup B$}\\
&= (M_n\setminus \bigcup_{i=0}^{n-1} M_i) \cup \bigcup_{i=0}^{n-1} (M_i\setminus \bigcup_{j=0}^{j-1}M_j) \quad\textit{using IH}\\
&= \bigcup_{i=0}^{n} (M_i\setminus \bigcup_{j=0}^{j-1}M_j).
\end{align}$$
