Did I find the right expression for the regular language for this FSA? I have the following FSA, and the regular language that I found for it:

Is this language correct? It doesn't match the solution in the book, but my teacher says there can be multiple equally correct languages. The book's solution looks like this: $(0\cup1)1^{*}00^{*}(11^{*}00^{*})^{*}$. Are these two languages equal?
 A: Yes, they are. You can simplify both to $$\Sigma^+\mathtt{0}$$
for $\Sigma = \{\mathtt{0},\mathtt{1}\}$, that is, $$(\mathtt{0}\cup\mathtt{1})(\mathtt{0}\cup\mathtt{1})^*\mathtt{0}$$
The crucial part is to observe that, after you cleared the initial state (i.e. read the first symbol), whatever you do, and wherever you are, after reading symbol $\mathtt{0}$ you get to the accepting state.
The difference between your solution and the solution from your book is that


*

*you split the final loop into two cases: either you read $\mathtt{11}^*\mathtt{0}$ or a bunch of $\mathtt{0}$'s;

*your book joins them saying that after reading $\mathtt{11}^*\mathtt{0}$ you are to to read all the $\mathtt{0}$'s left.


These are equivalent because both are in a loop, i.e. $B^*(ABB^*)^* = (AB \cup B)^*$.
I hope this helps $\ddot\smile$
A: Here is one calculation (inspired by dtldarek's comment):
Let $S_k$, $k=0,1,2$ represent the states above.
We have the equations $S_1 = (0|1) \big | S_1 1 \big | S_2 1$,
$S_2 = S_1 0 \big | S_2 0 $.
Solving for $S_2$ gives $S_2 = S_1 0 0^*$, then
$S_1 = (0|1) \big | S_1 1 \big | S_1 0 0^* 1$, which reduces to
$S_1 = (0|1) (1 | 0 0^* 1)^*$, and so
$S_2 = (0|1) (1 | 0 0^* 1)^* 0 0^*$.
Since $0 0^* = 0^* 0$, this gives $S_2 = (0|1) (1 | 0 0^* 1)^* 0^* 0$.
Note that $(1 | 0 0^* 1)^* = ((0|1)^* 1) \big | \epsilon$, that is, the empty string and all strings ending in $1$. Then we have
$(1 | 0 0^* 1)^* 0^* = (0|1)^*$, which gives $S_2 = (0|1)(0|1)^* 0$
Now consider $R=(0|1) 1^* 0 0^* (1 1^* 0 0^*)^* = (0|1) 1^* 0^* 0 (1 1^* 0^* 0)^* = (0|1)  (1^* 0^* 0 1)^* 1^* 0^* 0 $.
We have $(1^* 0^* 0 1)^* = ((0|1)^* 01 )\big | \epsilon$, that is, the empty string and all strings ending in $01$. Note that $1^*0^*$ describes the set of strings that have no $0 \to 1$ transition, and so we have
$(1^* 0^* 0 1)^* 1^* 0^* = (0|1)^*$ (either a string has a longest substring ending in $01$ or not). Hence $R = (0|1) (0|1)^* 0 = S_2$.
