Step 1: Compute successive powers of 7 modulo 5 directly: $7^1=7\equiv2\pmod5$; $7^2=49\equiv4\pmod5$; $7^3=343\equiv3\pmod5$; $7^4=2401\equiv1\pmod5$.
Step 2: The remainders repeat with cycle length 4: $2,4,3,1,2,4,3,1,\ldots$
Step 3: Divide the exponent 100 by the cycle length 4: $100=4\times25$ exactly, remainder 0, meaning $7^{100}$ sits at the 4th (last) position of the cycle.
Step 4: The 4th position value in the cycle is $1$.
\[\boxed{1}\]