Sum of unit and nilpotent element in a noncommutative ring. Something similar is asked here but it is not exactly the same thing.

An element $a$ of a ring $R$ is called nilpotent if $a^n=0$ for an $n \in \mathbb{N}$.  Show that if $u$ is a unit and $a$ is nilpotent in $R$, then $u+a$ is a unit.

My proof:

Let $a^n=0$.  The equality $$(u+a)^na^{n-1}=u^na^{n-1} \tag{1}$$  results from expanding $(u+a)^n$ together with the property that $a$ is nilpotent.  

Let $u_1$ be the left inverse of $u$.  Then by multiplying both sides of (1) on the left by $u_1^n$, the resulting equality is $u_1^n(u+a)^na^{n-1}=a^{n-1}$.  This is where I am nervous.  I thought that you could then write $u_1^n(u+a)^n=1$ by the property of the multiplicative identity in an associative ring with unity and setting the coefficents equal to each other, but I feel like this looks like applying a cancellation to the $a^{n-1}$ on both sides, which I do not mean to be doing.  

The last part is to rewrite the resulting equality as $(u_1^n(u+a)^{n-1})(u+a)=1$ to show the existence of a left inverse of $u+a$ and then do a similar thing starting from (1) to show the existence of a right inverse.

Is there an error with this?
This fails, for instance, 
$$\begin{pmatrix}0&amp;-1\\1&amp;0\end{pmatrix}+\begin{pmatrix}0&amp;1\\0&amp;0\end{pmatrix}=\begin{pmatrix}0&amp;0\\1&amp;0\end{pmatrix}$$
is singular.
In a ring $R$ the nilpotent elements constitutes an ideal which is called nilradical, $N(R)$ . 

$N(R)$ is the intersection af all prime ideals; so if $u$ is a unit and $a$ is nilpotent $a+u$ must be a  unit. 

Otherwise there would be a maximal ideal $M$ with $a+u \in M$. But a maximal ideal is prime, and so $$a \in N(R) \subseteq M$$ and so $$u = (u+a) - a  \in M$$ which is impossible because $u$ is a unit.