Does $5$ to the power of a positive integer $> 1$ always end in $25?$

No
Yes

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Prove that $\color{#D61F06}{5^n\equiv25\mod 100}$ , $\color{#D61F06}{\forall n\ge 2}$ , $\color{#D61F06}{n\in \mathbb{N}}$

Proof: Suppose that $\color{#D61F06}{n\in \mathbb{N}}$ . For $\color{#D61F06}{n=2}$ , we have \color{#D61F06}{5^2\mod 100 = 25\tag*{}}

or,

\color{#D61F06}{5^2\equiv 25 \mod 100\tag*{}}

so the base case holds.

Now, suppose that the statement holds true for some $\color{#CEBB00}{ n=k}$ . This allows us to write

\color{#CEBB00}{5^k\equiv 25 \mod 100 \tag{1}}

We are required to prove that the statement holds true for $\color{#CEBB00}{n=k+1}$

Multiplying both sides of equation $(1)$ by $5$ gives…

\color{#3D99F6}{\begin{aligned}5^k&\equiv 25\mod100\\5\cdot5^k&\equiv125\mod100\\5^{k+1}&\equiv\boxed{25}\mod100\end{aligned}}\tag*{}

Thus the statement holds true for $\color{#3D99F6}{n=k+1}$ , and hence the proof via mathematical induction.