square root of two is 1.41..., can we evaluate infinite(th) root of 2? I was just thinking about square roots. and then a thought struck me.
since square root of $2 = 2^{1/2} = 1.414..$
      cube root of $2 = 2^{1/3} = 1.259$
then  $2^{1/\infty} = 2^{0} = 1$
is this true that whatever number we take, $n^{1/\infty}$ root will be 1?
please correct me if i am wrong.
We tend to avoid treating $\infty$ as a number, because we lose a lot of nice rules about numbers when we do so.

What we do talk about is "limits." So we can look at the sequence of numbers:

$$2^{1/2},2^{1/3},2^{1/4},\dots$$

and we say that this sequence "has limit $1$" or that it "converges to $1$." Most students don't learn explicitly about limits until they get to calculus.

Most likely, the first place where you implicitly encountered a limit (without it being called such) is in repeated decimals.

For example, we write: $$\frac{1}{3}=0.3333\dots$$

This doesn't mean that $\frac{1}{3}=0.3$, nor does it mean that $\frac{1}{3}=0.33$. It means, rather, that $\frac{1}{3}$ is the "limit" of the sequence of numbers $$0.3,0.33,0.333,0.3333,\dots,$$

It took mathematicians a remarkably long time to formalize this notion of limit - it wasn't until the 19th century that they did so, even though they were implicitly using limits for much longer.
Yes, infiniteth root of $2$ is $1$.

$$\lim_{x\to\pm\infty} \sqrt[x]{2}=1$$