Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that $a_n$ converges to $c$. Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that the sequence converges to $c$.
My first problem was to find some terms of the sequence to verify that point and show that converges and converges to that point using some convergence criterion.
Could someone help me through this problem?

Note that calculating some values gives:
$$\Large \eqalign{
  & {a_1} = a  \cr 
  & {a_2} = \sqrt c \sqrt a   \cr 
  & {a_3} = \sqrt c \root 4 \of c \root 4 \of a   \cr 
  & {a_4} = \sqrt c \root 4 \of c \root 8 \of c \root 8 \of a  \cr} $$
In general you can prove that
$$\Large {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +  \cdots+ \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^{n-1}}}}}}$$
Since on has
$$\Large\mathop {\lim }\limits_{n \to \infty } \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +  \cdots \frac{1}{{{2^{n - 1}}}} = 1$$
and 
$$\Large\mathop {\lim }\limits_{n \to \infty } \root n \of a  = 1$$
for any real $a>0$, it is immeadiate that
$$\Large\lim {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +  \cdots \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^n}}}}} = {c^1} = c$$
