Number of zeros at the end of $99^n + 1$ is decided by the number of $9’s$ at the end of $99^n .$
For example, if there are $2,$ $9’s$ at the end of $99^n$ then $99^n + 1 = *****99 + 1 = *******00$
and if there are $3,$ $9’s$ at the end of $99^n$ then $99^n + 1 = *****999 + 1 = *******000$
Now, we have to decide how many $9’s$ are at the end of $99^n$ when $n$ is odd and $>1.$
$99^n$ has unit digit as $9$ when $n$ is odd and $>1$ because when we multiply $99$ even number of times, we get $1$ as unit digit and again multiply by $99$ will give $9$ as unit digit.
So, it confirms that $99^n$ ends with $9.$
Now, to get last $2$ digits, we have to do $mod\; 100$ operation on $99^n$.
$99^n = (100-1)^n = 100^n - n*(100)^{n-1}+.......-(n*(n-1)/2)*100^2+n*100-1$
Here, in the binomial expansion of $99^n$, all terms have $100$ as a multiple except $“-1”.$ So, all terms will give remainder as $0$ except $-1$ when divided by $100.$
So, $99^n\; mod\; 100= -1 \; mod\; 100 = (100-1) = 99$
So, it confirms that $99^n$ ends with $99.$
Now, to get last $3$ digits, we have to do $mod\; 1000$ operation on $99^n$.
when $n$ is odd and $>1,$ all the terms in the binomial expansion of $99^n$ have $1000$ as multiple except $n*100-1$ and binomial coefficients are always integers. So, all terms will give remainder as $0$ except $100n-1$ when divided by $1000.$
So, $99^n\; mod\; 1000= 100n-1$
Since, $100n-1\; mod\;1000$ will give $999$ when $n=10,20,30,…$ i.e. multiple of $10$ but here $n$ is odd, it means we can’t get last $3$ digits as $999.$
So, $99^n$ always ends with $99$ but not $999$ when $n$ is odd and $>1.$ So, number $99^n + 1$ must be ended with $2$ zeros.