Consider a bouquet of $n$ circles, centered at $(1,0), (2,0),...,(n,0)$ $$ W\_n= \bigcup\_1^nC\_r$$ with $$C\_r:(x-r)^2+y^2=r^2$$

I want to compute the fundamental group of this space.

> 
> $W\_n$ is path connected. Moreover I know
> $\pi\_1(W\_1)=\pi\_1(C\_1)=\mathbb{Z}.$ Seeing $W\_n= W\_{n-1}\cup C\_n$ we
> have both $W\_{n-1}$ and $C\_n$ are path connected and open. Moreover
> $W\_{n-1} \cap C\_n $ is just the origin which has trivial fundamental
> group. Hence
> $$\pi(W\_n)=\pi(W\_{n-1})\*\pi(C\_1)=\pi(W\_{n-1})\*\mathbb{Z}=...=\pi\_1(W\_1)\*\mathbb{Z}...\*\mathbb{Z}=$$$$\pi(C\_1)\*\mathbb{Z}\*...\*\mathbb{Z}=\mathbb{Z}\*...\*\mathbb{Z}$$
> so it is the $n$-times free product of $\mathbb{Z}$ with itself.
> 
> 
> 

Does it work? First I want to know if my solution is correct and only secondly a different one thanks.