No, in the end the private exponent $d$ is just a number within $0..N$ where $N$ is the modulus. It depends on $N$ what the chance is that the first bit is one, but in more likely to be valued $0$ than $1$ (given that it is well distributed, you would expect it to be $0$ around $\frac23$ of the time). If you generate enough private keys you'll even see private exponents that start with a few leading bytes that are completely zero'ed out (!). That's not a problem cryptographically speaking, it just shows that the entire key space is utilized.
So the private exponent $d$ doesn't always start with a $1$ when you consider the most significant bit of the modulus. The fact that the modulus is strictly less than $2^{keylength}$ does matter ever so slightly though. But as RSA requires exponential sized keys to achieve a certain security level, that should not worry you overly much (1 partial bit out of 1024 for the absolute minimum key size is completely negligible).
I'm assuming a calculation with a preset/precalculated RSA public exponent, which is most often used in practice. If PKCS#1 conformation calculations are used then the most significant bit of $d$ is always $0$.