Does the assumption $0^0=1$ ever lead to a contradiction or conflict with another useful assumption? There are some places where the assumption $0^0=1$ is formally useful.  For example when expressing polynomials
$$f\left[x\right]=\sum_{i=0}^{n}a_i x^i.$$
That leads me to question whether $0^0=1$ can be assumed in all cases.
Does the assumption $0^0=1$ ever lead to a contradiction or conflict with another useful assumption?
 A: Not sure if this is what exactly you're looking for, but I think if nothing else it's notable. 
Well, if we assume $0^0=1$, then in limits we should be able to use that whenever we evaluate by substitution (assuming continuity). As it happens, taking this idea from a pair of videos by blackpenredpen,
$$\lim_{x\to 0^+} x^x = 1 \;\;\;\;\; \text{but} \;\;\;\;\; \lim_{x\to 0^+} x^{\frac{1}{\ln(3x)}} = e$$
(Videos for the first limit and second limit are linked.)
The core idea for the first limit: we do some manipulations and utilize the continuity of the function on the positive reals, and then use L'Hopital's rule:
$$\begin{align}
\lim_{x\to 0^+} x^x &= \lim_{x\to 0^+} e^{x \ln(x)} \\
&= \exp \left(\lim_{x\to 0^+} x \ln(x) \right)\\
&= \exp \left(\lim_{x\to 0^+} \frac{\ln(x)}{1/x} \right)\\
&= \exp \left(\lim_{x\to 0^+} \frac{1/x}{-1/x^2} \right)\\
&= \exp \left(\lim_{x\to 0^+} -x \right) \\
&= \exp(0) \\
&=1
\end{align}$$
For the second limit, similar manipulations yield
$$\begin{align}
\lim_{x\to 0^+} x^{\frac{1}{\ln(3x)}} &= \exp \left( \lim_{x\to 0^+} \frac{\ln(x)}{\ln(3x)} \right)\\
&= \exp \left( \lim_{x\to 0^+} \frac{1/x}{1/(3x) \cdot 3} \right)\\
&= \exp \left( \lim_{x\to 0^+} 1 \right)\\
&= e
\end{align}$$
Notice however that naive substitution at the start for both yield a $0^0$ case. (In the second limit, $1/\ln(3x) \xrightarrow{x \to 0^+} 0$.)
