Question: Sorry for my bad english.

Automata: 

[![enter image description here](https://i.stack.imgur.com/x1TOE.png)](https://i.stack.imgur.com/x1TOE.png)

My answer:
$$L = \{0^n, 0^n1X \mid n = 1,2,3\dots\}$$

X is any strings that is not empty.

Answer: 
You have not accounted for the strings which begin with $1$. In your notation:
$$
\{1^n0^{m+1}|m,n\in \mathbb N\}\cup\{1^n0^{m+1}1X|m,n\in \mathbb N,X\text{ non-empty}\}
$$
where $\mathbb N=\{0,1,2.\dots\}$. In regex notation,
$$
(1^\*00^\*|1^\*00^\*1(0|1)(0|1)^\*)
$$
The automaton accepts any string containing at least one $0.$ In the notation I am used to, I would write $\{0,1\}^\*0\{0,1\}^\*$ as a regular expression for the language. Here $\{0,1\}^\*$ means any string of $0$'s and $1$'s, possibly empty. 

In the notation you are using the answer would be something like $$
L=\{1^n0, 1^n0X| n = 0,1,2,\dots\}$$ 
That is, a (possibly empty) string of $1$'s, followed by a $0$, followed by anything. If you have a notation (say Y) for a string that may be empty, then you can shorten it to $$
L=\{1^n0Y| n = 0,1,2,\dots\}$$