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
 A: While not a straightedge-and-compass procedure, and though providing only a fraction of the length sought, there's a way to "construct" $n^n/n!$ geometrically in the plane in just $n$ simple steps.
Identify a point $P_1$ on a unit circle. Anchor the segment of length $n$ at $P_1$, and then wrap the segment about the circle, reaching a point $P$. Starting at $P$, "peel back" curve $PP_1$ to determine its involute, $PP_2$. Then, again starting at $P$, peel back curve $PP_2$ to determine its involute, $PP_3$; continue until you reach involute $PP_n$ of curve $PP_{n-1}$.
Throughout the process, $|PP_i| = \frac{1}{n!} n^i$, so that $PP_n$ has the length promised.
(For length $n^n$ itself, you could take one more step to create $PP_{n+1}$ as the involute of curve $PP_n$. Then segment $P_n P_{n+1}$ has length $n^n/n!$, and you can just make $n!$ copies of that.)

The more-general result ---that starting with a circular arc $PP_1$ of length $\theta$ gives rise to involutes of length $\theta^i/i!$ for $\theta$ not-necessarily an integer--- is the foundation of Y. S. Chaikovsky's revelation of the geometry of the power series of sine and cosine. (See my Bloog post, "The Geometry of the Power Series for Trig Functions".)
A: 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 $i=n$.
