Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?

Let $R$ be a commutative ring with 1, we define

$$N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}$$

and

$$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$

Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?

I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.

This really is a problem in disguise: How did you derive the formula for the sum of the geometric series in year (something) at school??

$(a + 1)(a^{k-1} - a^{k-2} + \ldots 1) = 1-(-a)^n$

but as $a^n = 0$, we have (it does no matter whether $n$ is even or odd) that $(a+1)$ is invertible. Prove the following analogous problem, it may strengthen your understanding:

Let $A$ be a square matrix. If $A^2 = 0$, show that $I - A$ is invertible.

If $A^3 = 0$, show that $I - A$ is invertible.

Hence in general show that if $A^n = 0$ for some positive integer $n$, then $I -A$ is invertible.

Note the following:

$(1+a)(1-a) = 1-a^2$.

$(1-a^2)(1+a^2) = 1-a^4$

$(1-a^4)(1+a^4) = 1-a^8$

...

Thus, continuing in this way, we may find some $b_n$ such that $(1+a)b_n = 1-a^{2^n}$

For large enough $n$, this will give us $(1+a)b_n = 1$.

$N(R)$ is at least in some texts referred to as the nilradical of $R$. It is contained in all prime ideals (in fact, it is the intersection of all prime ideals, Atiyah, MacDonald prop. 1.8), so taking a nilpotent element $a$, since it is contained in all maximal ideals, $a+1$ is not in any maximal ideal. Then the ideal generated by $a+1$ must neccesarily be the whole ring, which means that it specifically generates 1 at some point.