Show that $L = \{1^n w 1^n | n > 0 \text{ and } w ∈ \{0,1\}^*\}$ is regular Show that the following language $L$ is regular by describing it using a regular expression.
  $$L = \{1^n w 1^n \mid n &gt; 0 \text{ and }w ∈ \{0,1\}^*\} $$


My (apparently incorrect) answer: 

Given $w ∈ \{0,1\}^*$ , we can rewrite $L$ as:  $1^n(0\cup1)^*1^n\ for\ n ≥ 0$

However, since the language specifies that the number of leading and trailing $1$s must be the same - this language is not​ ​regular​.

My justification:


For every regular language, there is (by definition) a finite automaton that accepts that language
The given language $L$ requires an equivalent count of leading and trailing $1$​s
Finite automata are not capable of maintaining a count.


Since​ ​no​ ​finite​ ​automata​ ​can​ ​accept​ ​the​ ​language​ $L$​,​ $​L$​ ​is​ ​not​ ​regular.

Can anyone explain to me why this language is regular?  I'm completely stumped.  I can see that it could be pumped using the Pumping Lemma (thereby proving regularity), but my brain just refuses to recognize that an FSM could accept this language.
My answer is just a hint. The following pattern should lead you to the right answer:

These strings belong to the language:
$w = 11 \Rightarrow 1^1\epsilon 1^1, n=1$
$w = 1011 \Rightarrow 1^1(01) 1^1, n=1$ 
$w = 11011 \Rightarrow 1^1(101) 1^1, n=1$ or $1^2(101) 1^2, n=2$ 
$w = 1001011 \Rightarrow 1^1(00101) 1^1, n=1$
$w = 11111 \Rightarrow 1^1(111) 1^1, n=1$ or $1^2(1) 1^2, n=2$ 
$w = 101010011 \Rightarrow 1^1 (0101001) 1^1, n=1$
$\cdots$
Finally, if $w=1^nw 1^n$ then it can be written as $1v1$, where $w,v \in (0+1)^*$. 

These strings DO NOT belong to the language:
$w=0$
$w=01$
$w=100$
$w=0000$
$w=0001$
$w=1000010$
$\cdots$

Yo may want to look at the first and the last symbols of each string.