Let $\Delta$ be infinitesimal, $\hat{e}=(1+\Delta)^{\frac{1}{\Delta}}$, and $log_{\hat{e}}(a)=\hat{ln}(a)$.
$$a^{\Delta}=({\hat{e}^{\Delta}})^{\hat{ln}(a)}=(1+\Delta)^{\hat{ln}(a)}$$
Consider $\frac{a^{\Delta}-1}{\Delta}$
$$\frac{a^{\Delta}-1}{\Delta}=\frac{a^{\Delta}-1}{log_a((a^{\Delta}-1)+1)}=\frac{1}{\frac{1}{a^{\Delta}-1}log_a((a^{\Delta}-1)+1)}=\frac{1}{log_a\left(({(a^{\Delta}-1)+1)}^{\frac{1}{a^{\Delta}-1}}\right)}$$
Using the definition of $\hat{e}$, which is one plus an infinitesimal brought to the power of the the infinitesimal's reciprocal, the last fraction becomes
$$\frac{1}{log_a(\hat{e})}=\hat{ln}(a)$$
Now you can solve for $a^{\Delta}$
$$\frac{a^{\Delta}-1}{\Delta}=\hat{ln}(a)$$ $$a^{\Delta}-1=\hat{ln}(a)\Delta$$ $$a^{\Delta}=1+\hat{ln}(a)\Delta$$
Substituting for $a^{\Delta}$ using the formula derived at the beginning of this post yields $$(1+\Delta)^{\hat{ln}(a)}=1+\hat{ln}(a)\Delta$$
since $\hat{ln}(a)$ can be any real number (and even any complex number) this becomes
$$(1+\Delta)^b=1+b\Delta$$
However, this is clearly false because
$$(1+\Delta)^2=1+2\Delta+\Delta^2$$
$$(1+\Delta)^3=1+3\Delta+3\Delta^2+\Delta^3$$ etc.
I would like to note that none of this is in terms of limits. Everything I did was exact even on the infinitesimal scale. The fact that the limits as $\Delta$ approaches $0$ of the right-hand sides equal the limits as $\Delta$ approaches $0$ of the left-hand sides doesn't resolve the problem.
I chose the denotations $\hat{e}$ and $\hat{ln}$ because $e=\lim_{x\to 0} ((1+x)^{\frac{1}{x}})$ so there is an infinitesimal difference between them that can be derived using binomial theorem or the Taylor series for $(1+x)^{1 \over x}$.
I would be exceedingly grateful if someone could shed light on this problem.
I think " ..the last fraction becomes" is your error, since you had $\hat e$ defined with one infinitesimal ($\Delta$) and "the last fraction" with a different one ($a^\Delta-1$). If the infinitesimals are different, you can't make the substitution.