This might appear like a unusual question, but I'd like to know how you can assume that $n^0 = 1$.
If $y$ is the sum of $n$ multiplied by itself $x-1$ times, then the first power of a number is just the same number with no multiplication, and the $0$th power is $y = n * n^{-1}$ (or a division of $n$ by itself), so $y = 1$.
But the natural logarithm also states that $y = \ln n^x$, so then $y = x \ln n$ and since $x = 0$, $y = 0 \ln n = 0$.
So shouldn't the $0$th power be $0$?
First i'd like to thank everyone who took time to answer, second here is what I Conclude :
It was just a simple basic rule to apply the logarithm rule of a quotient even if the le logarithm expression hold the value 0 inside such as $ln(1)$ = 0 if we divide anything by ln (1) it first apprears to me that we end up having a division by 0 so an an inequality where in fact the priority of operations put us to a substraction of both logarith instead
I think the confusion is coming from the fact that you only applied $\ln$ to the right side of the equation.
Since $y = n^x$, that means the correct way to apply the logarithm would be $$\ln y = \ln n^x$$
And since $\ln n^x = x \ln n$, if $x = 0$, that means $x \ln n = 0$, which makes $\ln y = 0$, or $y = 1$. Hope that helps!