A confusion about the formula for transistor Vbe Ic characterisitcs
The approximation is good enough because \$V_{be}\$ is a lot greater than \$V_t\$ (thermal voltage at a room temperature of \$27^o\$C).

\$V_t\approx 26\text{mV}\$ at \$27^o\$C.

Even for a value very close to zero, the approximation works well because the exponential term grows much larger than 1. 

Say \$V_{be}=100\text{mV}\$ (which is very close to zero), the exponential term becomes, \$e^{\frac{V_{be}}{V_t}}=46.81\$. This makes the 1 negligible because you could say 46.81 >> 1. Would it make much of a difference 46.81 (approx) vs 45.81(actual)? what is \$V_{be}\$ were 120mV? 101(approx) vs 100(actual)? If you're trying this in a lab, chances are that \$V_{be}\$ won't be exactly zero.

Obviously, the original equation, \$I_c=I_s(e^{\frac{V_{be}}{V_t}}-1)\$ will always work and will give you the exact result (If \$V_{be}\$ were exactly zero, this is the one you want to use because the exponential term no longer dominates). 

The approximation is valid when \$V_{be}\$ is such that the exponential term dominates, which is the case even for even a minuscule value of \$V_{be}\$, i.e 75mV.