Since $N(s)$ and $T(s)$ are normal for all $s$, we have
$\langle T, N \rangle = 0$
for all $s$. Thus
$0 = \langle T, N \rangle' = \langle T', N \rangle + \langle T, N' \rangle$,
whence
$\langle T', N \rangle = -\langle T, N' \rangle$;
by definition, $T' = \kappa N$; therefore
$\langle T, N' \rangle = -\kappa$.
Now since $\langle N, N' \rangle = 0$ we must have
$N' = -\kappa T$,
since $T$ and $N$ form an orthorormal basis for the tangent space at any point they are defined.
QED!
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Neither your answer nor the one that was given to you is correct. Note that\begin{align}s(t)&=\int_0^t\sqrt2e^t\,\mathrm dt\\&=\sqrt2\left(e^t-1\right).\end{align}So, you should take$$t=\log\left(1+\frac s{\sqrt2}\right).$$With this choice, $s=0\iff t=0$ as it should be, which fails to happen with any of those two solutions.
Best Answer
Yes, provided your curve has nonzero tangent vector at all points.
Suppose your curve is $\alpha: [a, b] \to \Bbb R^2$. For $t \in [a, b]$, define $$ q(t) = \int_a^t \| a'(s) \| ds. $$ You can see that $q(t)$ represents "how long is $\alpha$ from $a$ up to $t$".
What can you say about the function $q$?
$q(a) = 0$.
$q'(t) = \| \alpha'(t) \| > 0$ for every $t \in (a, b)$, by the fundamental theorem of calculus.
Define $L = q(b)$ to be the length of the whole curve.
Now: $q: [a, b] \to [0, L]$ is an increasing continuous function onto its codomain; hence it has an inverse function $u: [0, L] \to [a, b]$. We may not be able to easily write down the inverse, but it's there. And the derivative of $u$ at a point is (by the inverse function theorem) given by: $$ u'(t) = \frac{1}{q'(q^{-1}(t))} = \frac{1}{q'(u(t))} = \frac{1}{\|a'(u(t))\|}. $$
Hold that thought.
Now let $$ \beta: [0, L] \to \Bbb R^2 : t \mapsto \alpha(u(t)). $$ Clearly $\beta$ traverses the same path as $\alpha$. But what's $\beta'(t)$? It is, by the chain rule, \begin{align} \beta'(t) &= \alpha'(u(t)) \cdot u'(t)\\ &= \alpha'(u(t)) \cdot \frac{1}{\| \alpha'(u(t))\|}, \end{align} which is a unit vector. QED.