Plane Geometry $n^n$ construction Given a line segment of length $n$. how to construct $$n^n, n \in N$$ using straight edge and compass
Example:- given line segment's of length a and b. i can construct $$\sqrt{\mathbf a^2 + \mathbf b^2}$$ by drawing line segment a with straight edge and drawing line segment b perpendicular to line segment a at one end. then joining the other end's of line segment a and b we get the desired result. 

Regards,
vishal

Here's a path to $n^n$ exactly, via legitimate straightedge-and-compass constructions. (Hereafter, "construct" means "construct with straightedge and compass".)
If you have a segment of length $n$, then you can construct a segment of length $1$, and therefore also a right triangle $\triangle P_0OP_1$ with legs $|OP_i| = n^i$. For each $1 \le i \le n-1$, construct line $p_i$ perpendicular to $P_{i-1}P_i$ at $P_i$, and let $P_{i+1}$ be the point where $p_i$ meets ray $\overrightarrow{P_{i-1}O}$; then $\triangle P_iOP_{i+1}$ is a right triangle (with right angle at $O$) similar to $\triangle P_{i-1}OP_{i}$ (and thus is also ultimately similar to $\triangle P_0OP_1$).
The resulting chain of similar triangles yields this chain of proportions:
$$n = \frac{|OP_1|}{|OP_0|} = \frac{|OP_2|}{|OP_1|} = \cdots = \frac{|OP_{i+1}|}{|OP_i|}= \cdots = \frac{|OP_n|}{|OP_{n-1}|}$$
whence $|OP_i| = n^i$ for all $i$, including $i=n$.
