I am facing some difficulties proving the reverse operation on N elements.
Let Σ be an alphabet, given a word x ∈ Σ∗ and rev(x) the reverse of x. I want to prove that
for n >= 1 and for X1,X2,...,Xn ∈ Σ∗
rev (X1,X2,...,Xn) = rev(Xn),...,rev(X1)
I would appriciate your assistance.
You can prove it by (strong) induction on $n$. The base case $n=1$ is trivial, since it just says that $\operatorname{rev}(x_1)=\operatorname{rev}(x_1)$. For the induction step you need to show that if $n>1$, and
$$\operatorname{rev}(x_1,\ldots,x_m)=\operatorname{rev}(x_m),\ldots,\operatorname{rev}(x_1)$$
for arbitrary $x_1,\ldots,x_m\in\Sigma^*$ whenever $1\le m<n$, then
$$\operatorname{rev}(x_1,\ldots,x_n)=\operatorname{rev}(x_n),\ldots,\operatorname{rev}(x_1)$$
for arbitrary $x_1,\ldots,x_n\in\Sigma^*$.
If $n>2$, you can let $y=x_1,x_2,\ldots,x_{n-1}$; then $x_1,\ldots,x_n=y,x_n$, so
$$\operatorname{rev}(x_1,\ldots,x_n)=\operatorname{rev}(y,x_n)=\operatorname{rev}(x_n),\operatorname{rev}(y)$$
by the induction hypothesis with $m=2$, and $\operatorname{rev}=\operatorname{rev}(x_{n-1},\ldots,x_1)$ by the induction hypothesis with $m=n-1$, so $\operatorname{rev}(x_1,\ldots,x_n)=\operatorname{rev}(x_n),\operatorname{rev}(x_{n-1}),\ldots,\operatorname{rev}(x_1)$, as desired.
Unfortunately, this still leaves the $n=2$ case to be dealt with: you have to show that for any $x_1,x_2\in\Sigma^*$, $\operatorname{rev}(x_1,x_2)=\operatorname{rev}(x_2),\operatorname{rev}(x_1)$. You can do this by writing out $x_1$ and $x_2$ in terms of elements of $\Sigma$ (e.g., $x_1=a_1a_2\ldots a_r$ and $x_2=b_1b_2\ldots b_s$) and expressing $\operatorname{rev}(x_1,x_2)$, $\operatorname{rev}(x_1)$, and $\operatorname{rev}(x_2)$ directly in terms of elements of $\Sigma$.
A more formal approach requires defining the reversal operation more formally, first on strings of elements of $\Sigma$ (i.e., on members of $\Sigma^*$), then on strings of elements of $\Sigma^*$ (which is what you need here). The former is done in this question.