Showing that the entropy of i.i.d. random variables is the sum of entropies
Your attempt at the proof does not take into account any natural order on $\Sigma^n$. While it may still work, it seems difficult.

A much easier proof can be given by induction over $n$. The base case is trivial ($H(X)=H(Y)$) for $X=Y$.
Then, denote $X=X_1\cdots X_n$ and $X'=X_1\cdots X_{n-1}$, so you have

$$H(X=w)=-\sum_{w\in\Sigma^{n-1}}\sum_{\sigma\in \Sigma}Pr(X=w\sigma)\log Pr(X=w\sigma)$$
$$=-\sum_{w\in\Sigma^{n-1}}\sum_{\sigma\in \Sigma}Pr(X'=w)Pr(Y=\sigma)\log Pr(X'=w)Pr(Y=\sigma)=$$
$$=-\sum_{w\in\Sigma^{n-1}}Pr(X'=w)\sum_{\sigma\in \Sigma}Pr(Y=\sigma)(\log Pr(X'=w)+\log Pr(Y=\sigma))=$$
$$=-\sum_{w\in\Sigma^{n-1}}Pr(X'=w)\left(\sum_{\sigma\in \Sigma}Pr(Y=\sigma)\log Pr(X'=w)+\sum_{\sigma\in \Sigma}Pr(Y=\sigma)\log Pr(Y=\sigma)\right)=$$
$$=-\sum_{w\in\Sigma^{n-1}}Pr(X'=w)\log Pr(X'=w)\sum_{\sigma\in \Sigma}Pr(Y=\sigma)+\sum_{w\in \Sigma^{n-1}}Pr(X'=w)\sum_{\sigma\in \Sigma}Pr(Y=\sigma)\log Pr(Y=\sigma)=$$
$$=H(X')+H(Y)=(n-1)H(Y)+H(Y)=nH(Y)$$