Why does it seem as if I can apply the Pumping lemma to a language that is regular? We learn about the Pumping Lemma at the class and I tried to make few examples to understand it...

There I make this example:
Let's say: $L=\{w\in L|w=(0+1)^*1\}$ - i.e. - L is the language of all the words that finish with $1$. (The language is regular of course).  

Now, I can take the word: $w=0^{n-1} 1$. $w\in L$ and $|w|\ge n$.
I can take: $u=0^{n-1},v=1,z=\varepsilon$.
$|uv|\le n$ and $|v|&gt;0$, but:
$uv^0z\notin L$ because $uv^0z$ ends with $0$...

What I miss here?
I'd like to understand more where is my mistake...
The pumping lemma says that, for every regular language there exists some pumping length&nbsp;$p$* such that blah blah blah.

You made two mistakes. First, you tried to choose&nbsp;$p$ based on a specific word, rather than based on the language as a whole. Second, you chose the wrong value of&nbsp;$p$. In particular, for the language you've given, $p=1$ has the property required of the pumping lemma: for every word&nbsp;$w$ in that language, you can write $w=\epsilon y z$ with $|y|=1$, and you'll find that $\epsilon y^n z$ is in the language for all&nbsp;$n\geq 0$.