Suppose Arsenii's claim is correct and by changing the last digit of number $N$ he obtained $13^k$, where $k$ is a positive integer. Clearly, he had to change the last digit, because it is $5$. Since $2^{100} \equiv 1 \pmod{3}$, $3^{100} \equiv 0 \pmod{3}$ and $5^{100} \equiv 1 \pmod{3}$, the sums of their digits have the same remainders modulo $3$. The change of the last digit $5$ to digit $l$ means that we subtracted $2$ (or added $1$) and added $l$ modulo $3$. Thus, the new number (denoted by $M$) equals $l \equiv 0 \pmod{3}$. Since $13^k \equiv 1 \pmod{3}$, we could add one of the following digits: $1$, $4$ or $7$. Since the last digit of $13^k$ can be $1$, $3$, $7$ or $9$, we need to consider two cases.
Case 1. $13^k$ ends with $1$, it is possible in the case $k \equiv 0 \pmod{4}$, then the following is true modulo $8$: $M = 13^k = 13^{4j} = 169^{2j} \equiv 1 \pmod{8}$. Note that $M = N - 4$. $N \equiv 5^{100} = 25^{50} \equiv 1 \pmod{8} \Rightarrow M = N - 4 \equiv 5 \pmod{8}$, so we get a contradiction.
Case 2. $13^k$ ends with $7$, it is possible if $k \equiv 3 \pmod{4}$, then the following is true modulo $8$: $M = 13^k = 13^{4j+3} = 169^{2j} \cdot 13 \equiv 5 \pmod{8}$. Note that $M = N + 2 \Rightarrow M \equiv 3 \pmod{8}$ – contradiction.
These contradictions complete the proof.