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This is a part of a simple derivation in my textbook, but I'm really not seeing where it came from.

$\frac{\vec{E}\cdot d\vec{s}}{|d\vec{s}|} = E \cdot \vec{n}$ where n is the normal vector to $d\vec{s}$. However, shouldn't n be the unit vector rather than the normal vector?

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Maybe you are (reasonably) confusing the notion of normal vector with the notion of a normalized vector (dividing by its norm to make it have norm equal to one)?

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    No, isn't it supposed to be the unit vector (same direction)? – 2017-02-13
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    that is the normalized vector – 2017-02-13
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    @SakethMalyala Oh ok, but I know the difference between the two and the textbook says normal vector. – 2017-02-13
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Yes you are correct. $\frac{\vec{x}}{|\vec{x}|}$ is the unit vector in the direction $\vec{x}$, not normal to $\vec{x}$.

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    The book seems to consistently use it as normal vector. The topic is stoke's theorem. – 2017-02-13
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    Take the vector $(1,0)$ which has norm one, and so the resulting vector is the original, which is obviously not normal to $(1,0)$. So, something is off. – 2017-02-13
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    @Goldname: If it is about Stokes' Theorem, then this is even more weird. In Stokes' Theorem, the normal vector multiplies the curl of the field. The field, in the line integral, is multiplied by the tangent. – 2017-02-13