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Let $R$ be a ring (not necessarily commutative). Let $u\in R^\times$, (so $u$ is a unit), and let $a\in R$ be a nilpotent element. Show that if $ua=au$ then $u+a\in R^\times$.

I have that if $a$ is nilpotent, then $1-a$ is a unit. Then, I know that this means $\exists b\in R$ such that $ (1-a)b=1$. Additionally I have that $\exists c\in R$ such that $uc=1$. Then I was thinking maybe I could do some kind of algebraic manipulation like $uc=(1-a)b$ but that seems to only lead to dead ends.

Any hints or pointers in the right direction appreciated.

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    The title seems to have nothing to do wih the actual question... – 2017-01-20
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    @Arthur OOPS!! Fixed! – 2017-01-20
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    Again, the title does not have anything to do with the question. Moreover, the title claims something false. In $\mathbb Z$ the elements $2$ and $3$ commute and their sum is not a unit. – 2017-01-20
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    @MarianoSuárez-Álvarez Better? Titles are hard lol – 2017-01-20
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    $u + a = u (1+u^{-1} a)$ so I would try to show that $u^{-1} a$ is nilpotent. – 2017-01-20
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    Pointer : use the site's search feature first. There are at least two duplicates of this, and they show up at the very top of the most naive search. – 2017-01-20

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It suffices to show that $1+\frac{a}{u}$ is a unit. Since $\frac{a}{u}$ is still nilpotent, it suffices to prove that $1+a$ is a unit for nilpotent $a$, and this follows from $$\frac{1}{1+a}=1+a+a^2+\cdots +a^n$$