Imagine we have a language like $L$ with alphabet $\Sigma$ and the set of words of $L$ called $\Sigma^*$ ( notice that a word can have zero characters). We define $\delta^*$ recursively like this :
For each $a \in \Sigma$, $w \in \Sigma^*$ and $q \in Q$ :
$(1)$ $\delta^*(q,\lambda)=q$
$(2)$ $\delta^*(q,wa)=\delta(\delta^*(q,w),a)$
Now prove that for each two strings like $w$ and $v$, $\delta^*(q,wv)=\delta^*(\delta^*(q,w),v)$.
Note : what i tried was induction on $|v|$ but it didn't work.
For $|v|=0$ we have this: $\delta^*(q,wv)=\delta^*(q,w\lambda)=\delta^*(q,w)=\delta^*(\delta^*(q,w),\lambda)=\delta^*(\delta^*(q,w),v)$.
Assume that for $|v|=n-1$ its true that : $\delta^*(q,wv)=\delta^*(\delta^*(q,w),v)$.
Now i'm stuck on this : How to show that its true for $|v|=n$ ?
By induction on $|v|$.
If $v=\lambda$, then $\delta^*(q,wv)=\delta^*(q,w)=\delta^*(\delta^*(q,w),\lambda)=\delta^*(\delta^*(q,w),v)$.
Assume that for all $v$ such that $|v|=n-1$ the property is true.
Let $v$ be such that $|v|=n$, assume (without loss of generality) that $v=v'a$. Then: using (1) $$\delta^*(q,wv)=\delta^*(q,wv'a)=\delta(\delta^*(q,wv'),a)$$ using the induction hypothesis we know that: $$\delta^*(q,wv')=\delta^*(\delta^*(q,w),v')$$ thus $$ \delta^*(q,wv)=\delta(\delta^*(\delta^*(q,w),v'),a)$$ by applying (2) we obtain: $$ \delta^*(q,wv)=\delta^*(\delta^*(q,w),v'a)$$ thus $$\delta^*(q,wv)=\delta^*(\delta^*(q,w),v)$$