Prove .....

$99\equiv(-1)\pmod{100} \Rightarrow 99^{n}\equiv(-1)^{n}\pmod{100}.$But,whenever $n$ is an odd integer$, (-1)^{n}=(-1) \Rightarrow 99^{n}\equiv(-1)\equiv(99)\pmod{100}$ for all odd $n.$

This is similar to prove that

$(100-1)^n\equiv 99 \mod 100$

For $n$ is an odd number.

Noticed that every term of the expansion is divisible by $100$ except for the last term, $-1$. Hence we get

$(100-1)^n\equiv -1 \mod 100$

$(100-1)^n\equiv 99 \mod 100$

For this , let us consider two cases when $n$ is odd and when it is even ,

When $n$ is even :

$99$ could be expressed as $100-1$. So, we could write it as $(100-1)^{n}$. As in this case we have assumed $n$ to be even us let us write $n=2k$ . So we have $(100-1)^{2k}$ . Now , little bit expanding it , we have , $10000+1-200)^{k}$ . Now taking $p=10000+1$ and $q=-200$ , we have $(p+q)^{n}$ . Now let us prove one more result , for any $(100+1)^{x}$ , the last two digits would be $01$ . By expanding by binomial theory , $(100+1)^{x} = 100^{x}+a_{1}100^{x-1} \times 1 + . . . . a_{y}100 \times 1^{x-1} + 1^{x}$ So, expanding it will have $x+1$ terms. It is clear that the first $x$ terms would have $00$ in their last as they are multiples of some powers of $100$. But the last term will be 1. So , returning to our expansion we, would have , $p^{n} +a_{1}p^{n- 1} q^ {1} + . . . . . a_{y} p q^{n-1} + q ^{n}$ Now in this we have except $p^{n}$ all other terms would be a multiple of $100$ as $q$ is a motile of it . As we have earlier proove that the last term of $p^{n}$ would be 1 . And , as all other terms end in $00$ , therefore 3rd two digits of the whole computation would be $01$ . So we have thus profed that any power that is even would yield a result of 01 at the end when applied to $100-1$. Case 2 could be proved in a similar way and after proving it we would have proved the result so instead doing that long proof for case 2 also we could say that any odd number can be expressed $n=2k+1$ so just multiplying one more $99$ to any even power we would obtain the odd power. And multiplying the end two digits which are $01$ for any even power of 99 as we have proved , by 99 would give us 99 at the end. Hence , proved.

QED