TITLE: How to prove the following inequality without expansion
QUESTION [5 upvotes]: $M^k \le 2^r < M^{k+1}$
where $M>1 , k>0$ for some $r$.
It simply tells you that there exists a $2^r$ between $M^k$ and  $M^{k+1}$. for example:
if $M=3$, $k=1$ then
$$M^k = 3, \quad M^{k+1} = 9$$ 
and there exists $4$ and $8$ in between $3$ and $9$. i.e., $2^2$ and $2^3$

Edit: (T.B.)

Let $M \geq 2$ and $k \geq 1$ be integers. How can I prove that there exists an integer $r$ such that
  $$M^{k} \leq 2^r \lt M^{k+1}\quad?$$

REPLY [2 votes]: The simplest way to state this is using the natural logarithm : the inequality
$$
M^k \le 2^r <M^{k+1} 
$$
is equivalent to
$$
k \log M \le r \log 2 < (k+1) \log M
$$
and again equivalent to
$$
k \left( \frac{\log M}{\log 2} \right) \le r < (k+1) \left( \frac{ \log M}{ \log 2} \right).
$$
which clearly has a solution because this is asking if there is an integer in the interval $[ k(\log M / \log 2), (k+1) ( \log M / \log 2 ) )$. It is possible because $M \ge 2$, hence the fraction of logarithms is greater than $1$ and the interval has length bigger than $1$.
Hope that helps,