If $0^0 = 1$, then is it true that $0/0 = 1$? By Knuth, Concrete Mathematics (2nd ed.) page 162, it is convenient that 
$$0^0 = 1$$
Then, is it true that
$$0^1/0^1 = 0^{1-1}= 0^0 = 1$$ and we are free to exclude indeterminate statement of $0/0$ ?
 A: There's a good reason why it makes more sense to define $0^0$ as $1$ than it does to define $0/0$ as $1$. Notice that $0=\alpha\cdot 0$ for any real number $\alpha$. Substituting this in $\frac00=1$ would give $1=\frac00=\frac{\alpha\cdot 0}{0}=\alpha$, so this convention would not be consistent.
Conversely, the convention that $0^0=1$ is consistent, at least by this measure, because $0^{\alpha\cdot 0}=(0^0)^\alpha=1^\alpha=1$, and $(\alpha\cdot 0)^0=\alpha^0\cdot0^0=1\cdot 1=1$.
A: No, the argument doesn't pass through.
You cannot define $0^{-1}$, so using $0^{1-1}=0^1\cdot0^{-1}$ is invalid.
A: Assuming that $0^0=1$ is wrong because, if you take natural logarithms both sides, this will mean $$0=\ln (1) =\ln(0^0)=0 \cdot \ln (0)=0 \cdot \lim_{x \to 0^{+}} \ln (x)=0 \cdot (-\infty) $$ which is a contradiction, because $0 \cdot (-\infty)$ is an indeterminate (an operation that is meaningless.)
Similarly, suppose $\frac{0}{0}=1$ as evaluated by yourself, then that will mean
$$\infty -\infty = \ln \left( \frac{0}{0}\right)=\lim_{x \to 0} \ln (x) -\ln(x)=\ln(1)=0$$ which is also a contradiction.
Let me supply an example, we know that $0.0000001 \approx 0 $ and $0.01 \approx 0$, then $$\frac{0}{0} \approx \frac{0.01}{0.0000001}=100000>>1$$
Let me give another example, since the symbol $\infty$ means a biggest number, I can consider the number 120 to be my $\infty$ just like in a T-distribution table so that $$\infty-\infty=125-120=5 >0$$ This is the reason why these operations are called indeterminate.
