Is there a quicker way to get $ 207^{321} \mod 7 $ For $207^{321} \pmod{7},$ 
I got $$ 207^{321} = 207^{6\cdot 53+3}$$
and
$$207^{\Phi(7)} \equiv 207^6 \equiv 1 \pmod{7}$$ by Euler's Theorem.
Then 
$$207^3  \equiv 4^3 \equiv 1 \pmod{7} $$
Is there any simpler way? 
I'm also not sure about the format of module symbol.Should there be only one (mod 7) written on the right of the equation so as to avoid redundancy ?
I have also seen equation like this 26 mod 5=1,rather than $26\equiv 1 \mod{5}$. What's the difference?
 A: I don't know, your line of reasoning seemed pretty quick and easy to me.  I suppose you could observe that (modulo $7$)
$$
207^{321} \equiv (207 \bmod 7)^{(321 \bmod 3)} \equiv 4^0 = 1
$$
Is that simpler, by your lights?
The difference in notation, incidentally, is that when you write $26 \bmod 5 = 1$, the mod is treated as an operation—essentially the remainder left when you divide $26$ by $1$.  When you write $26 \equiv 1 \pmod 5$, you mean that $26$ and $1$ fall into the same equivalence class, modulo $5$.  The two formulations are equivalent under ordinary circumstances.
A: Note that $207^{321}=(207^{6})^{53} \cdot 207^3=1\cdot 4^3=1$ (mod. $7$).
A: You could use $a^p\equiv a\pmod p$ a Fermat"s little theorem derivative,then use exponent rules to show only the digital root in prime base $p$ of any exponent matters to this calculation. So as $321_{10}=636_{7}$  which has digital root $3$ we can decrease the exponent to $3$ we then apply the fact that $(px+y)^n$ has only $1$ term that doesn't include a factor of $p$ so we can calculate $y^3$ in this case.  Your way uses much of the same theory but is less verbose.
