The shannon entropy of a random variable $Y$ (with possible outcomes $\Sigma=\{\sigma_{1},...,\sigma_{k}\}$) is given by
$H(Y)=-\sum\limits_{i=1}^{k}P(Y=\sigma_{i})\;\log(P(Y=\sigma_{i}))$.
For a second random variable $X=X_{1}X_{2}...X_{n}$, where all $X_{i}$'s are independent and equally distributed (each $X_{i}$ is a copy of the same random variable $Y$), the following equation is known to be true:
$H(X)=n\cdot H(Y)$
I want to prove this simple equation, where the outcomes from $Y$ are interpreted as symbols from an alphabet $\Sigma$ and therefore $X$ is the random variable for strings of length $n$ (based on the distribution of $Y$).
It is easy to see, that
$P(X=w)=P(X=w_{1}...w_{n})=P(X_{1}=w_{1})\;\cdot\;...\;\cdot\; P(X_{n}=w_{n})=\prod\limits_{i=1}^{n}P(Y=w_{i})$
... but my approach to prove $H(X)=n\cdot H(Y)$ seems to be in a dead point
(every word $w$ has the form $w=w_{1}...w_{n}$ with $w_{i}\in\Sigma$ and let $\large |w|_{\sigma_{i}}$ be the number of occurrences of $\sigma_{i}$ in $w$):
$H(X)=-\sum\limits_{w\in\Sigma^{n}}P(X=w)\;\log(P(X=w))$
$=-\sum\limits_{w\in\Sigma^{n}}\left(\prod\limits_{i=1}^{n}P(Y=w_{i})\right)\;\log\left(\prod\limits_{i=1}^{n}P(Y=w_{i})\right)$
$=-\sum\limits_{w\in\Sigma^{n}}\left(\prod\limits_{i=1}^{n}P(Y=w_{i})\right)\left(\sum\limits_{i=1}^{n}\log\left(P(Y=w_{i})\right)\right)$
$=-\sum\limits_{w\in\Sigma^{n}}\left(\prod\limits_{i=1}^{k}P(Y=\sigma_{i})^{\large |w|_{\sigma_{i}}}\right)\left(\sum\limits_{i=1}^{k}\large |w|_{\sigma_{i}}\normalsize \;\log\left(P(Y=\sigma_{i})\right)\right)$
So, I am able to change some indices from word length to the length of the alphabet, which is used in $H(Y)$. But what now? Any help?