It is easy to prove that if 1 is raised to any power of a rational number ( say $x$), then :
$(1)^x = (1)^{p/q}$
(because $x$ can be expressed as $p/q$ where $p$ and $q$ are integers except $q≠0$). And,
$(1)^{(p/q)} = (1)^p/(1)^q = 1/1 = 1$
But I am stuck in proving following :
a) $1^π = 1$
b) $1^e = 1$
c) $1^{\sqrt3} = 1$
Or in general, how to prove $1^{y}$ = 1 , where $y$ is an irrational number.
I would appreciate any help in proving above things .
You first need to understand what $1^x$ means when $x$ is irrational. If $x$ is rational, then it can be written as $x=p/q$ in reduced form. Then, for any real number $a$, the definition of $a^x$ is $$ \sqrt[q]{a^p}, $$ where $a^p=a\cdot a\cdots a$ when $p>0$, and $a^p=(1/a)\cdot (1/a)\cdots (1/a)$ when $p<0$ (both repeated $|p|$ times).
This does not work when $x$ is irrational, so what to we do? The idea is to approximate $x$ by rational numbers. For example, to compute $a^\pi$, you could look at the sequence of numbers $$ a^3,\;a^{3.1},\;a^{3.14},\;a^{3.141},\dots $$ Each of these makes sense since any terminating decimal can be written as a fraction. As the exponents get closer and closer to $\pi$, it turns out that the list of numbers will converge to some number. We call the number they converge to $a^{\pi}$. To make all this precise, you need to understand what the limit of a sequence means.
To prove that $1^{\pi}=1$, you just have to note that $1^{\pi}$ is defined as the limit of the sequence $1^{3},1^{3.1},1^{3.14},\dots$, and these numbers are all $1$ because of the rules you stated in your question. Since the limit of a sequence of $1$'s is just $1$, we conclude $1^\pi=1$. The same applies to any irrational number $x$; it is always true that $1^x=1$.