How can I prove this two formula $| M^{n} | = | M |^{n}$ I have no idea, how to prove this.


*

*$\left| M \cdot N \right| = \left| M \right| \cdot \left| N \right|$

*$\left| M^{n}  \right|  = \left| M \right|^{n} $
please help me. 
it would be so nice to see a answer.
thank you
EDIT
$$
2) M^{k}  := \left\{ \left(m_{1}, m_{2}, .... ,m_{k} \right) : m_{1}, m_{2}, .... ,m_{k} \in M          \right\} 
$$
I have to prove it with induction
Its about 2 Sets.
EDIT 2
$$
2) \\
M^{1}  := M \\
M \neq \left\{ \right\} \\
n \in \mathbb N 
$$
 A: Edit: Note that if $(m,n)\in\{m\}\times N$ for some $m\in M$, then $(m,n)\in M\times N$. On the other hand, if $(m,n)\in M\times N,$ then $(m,n)\in\{m\}\times N.$ Note moreover that the $\{m\}\times N$ are pairwise disjoint, meaning that if $m,m'\in M$ with $m\ne m'$, then $\{m\}\times N$ and $\{m'\}\times N$ have no elements in common. Thus, since $$M\times N=\bigcup_{m\in M}\{m\}\times N$$ and the union is disjoint, then  $$|M\times N|=\sum_{m\in M}\bigl|\{m\}\times N\bigr|.$$ You should be able to see an easy bijection $\{m\}\times N\to N$ for any $m\in M$, so $\bigl|\{m\}\times N\bigr|=|N|$ for each $m\in M$, whence $$|M\times N|=\sum_{m\in M}|N|=|M|\cdot|N|.$$
The other part is a fairly simple induction. The $k=1$ case is easy, since $M^1=M$. Note that $|M^{k+1}|=|M^k\times M|.$ Indeed, $M^2=M\times M=M^1\times M$, and if $k\ge 2$, define $f:M^{k+1}\to M^k\times M$ by $$(m_1,\dots,m_k,m_{k+1})\mapsto\bigl((m_1,\dots,m_k),m_{k+1}\bigr).$$ You can show this to be a bijection, whence your conclusion follows inductively from the first part, since if $|M^k|=|M|^k,$ then $$|M^{k+1}|=|M^k\times M|=|M^k|\cdot|M|=|M|^k\cdot|M|=|M|^{k+1}.$$
